NCERT Solutions · Class 11 Physics Motion in a Straight Line

A body counts as a "point object" only when its own size is too small to matter next to the distances involved in describing its motion. If turning the object's size or shape "on" or "off" does not change the path you'd draw for it, you may shrink it to a point and ignore everything about its internal structure — spin, orientation, tumbling. If the object's size, spin, or orientation is itself part of what is happening, it cannot be reduced to a point.Check each case against that test.(a) A railway carriage moving without jerks between two stations The carriage is a few metres long, but the distance between stations is kilometres — the ratio of carriage-length to path-length is negligible. "Without jerks" also means every part of the carriage moves together, with no internal rotation or relative motion between its parts that would need tracking separately. Nothing about the carriage's own size or shape changes how you'd describe its journey from one station to the other. This body can be treated as a point object.(b) A monkey sitting on top of a man cycling smoothly on a circular track Over one lap of a circular track, the path length (the track's circumference) is again far larger than the size of the man-and-monkey system. "Sitting" and "cycling smoothly" mean the monkey has no motion relative to the man — the two move as a single rigid body with no internal jerks to account for. So, exactly as in (a), the whole system's extent is negligible against the distance it travels. This body can be treated as a point object.(c) A spinning cricket ball that turns sharply on hitting the ground Here the sideways deviation of the ball after it bounces is caused directly by its spin — by rotation about its own axis. That rotation, and the ball's own size (which sets how the spin couples to the ground through friction), is not something you can discard: it is the very thing responsible for the sharp turn you are trying to describe. Shrinking the ball to a point would erase the effect you are asked to explain. This body cannot be treated as a point object.(d) A tumbling beaker that has slipped off a table edge "Tumbling" means the beaker is rotating end-over-end as it falls — its orientation is changing continuously and matters (for instance, to how and where it lands). Since the fall height is comparable to the beaker's own size, and the tumbling motion depends on that size and shape, the beaker's extent cannot be ignored. This body cannot be treated as a point object.Answer: The railway carriage (a) and the monkey-cyclist system (b) can be treated as point objects, since their own size is negligible compared to the distances they travel and neither shows any internal rotation. The cricket ball (c) and the tumbling beaker (d) cannot, because their spin/tumbling — which depends on their own size and shape — is essential to describing their motion.

On a position-time graph, everything you need is hidden in four features of each line: where it starts on the time axis, where it ends on the height axis, how steep it is, and where it crosses the other line. Read those four features one at a time and each blank fills itself in.The vertical axis is position \(\displaystyle x \), measured from the school \(\displaystyle O \), and the horizontal axis is time \(\displaystyle t \). A straight line on an \(\displaystyle x \)-\(\displaystyle t \) graph means the child is walking at a constant speed, because the slope of that line is\[v = \frac{\Delta x}{\Delta t} \]— the same slope everywhere on a straight segment, so "steeper" always means "faster," no matter where on the graph you measure it.(a) Who lives closer to school — compare where the lines end. The home of each child is the point where that child's line stops rising, i.e. its final height on the \(\displaystyle x \)-axis. That final height is exactly \(\displaystyle OP \) for A and \(\displaystyle OQ \) for B. In Fig. $\displaystyle 2.9$ A's line ends at a smaller \(\displaystyle x \)-value than B's line, so \(\displaystyle OP < OQ \). A's home is nearer the school. \(\displaystyle (a)\) A lives closer to the school than B.(b) Who leaves school earlier — compare where the lines start. A line that begins farther to the left on the \(\displaystyle t \)-axis represents a child who left the school (i.e., \(\displaystyle x = 0 \)) at an earlier clock time. A's line leaves the \(\displaystyle t \)-axis before B's does — B's line only lifts off \(\displaystyle x=0 \) later. A common mixup here is to look at who arrives first instead of who departs first; the question is about the start of the walk, which is the left-hand endpoint of each line, not the right-hand one. \(\displaystyle (b)\) A starts from the school earlier than B.(c) Who walks faster — compare the slopes, not the starting times. Speed is read from the slope \(\displaystyle \dfrac{\Delta x}{\Delta t} \), and slope is independent of when a line happens to begin. Even though A leaves first, B's line rises more steeply — B has the farther home \(\displaystyle Q \) to reach and leaves later, so to arrive on the same schedule as A, B must cover more ground in less available time, which forces a larger \(\displaystyle \dfrac{\Delta x}{\Delta t} \). The steeper line in the graph is B's. \(\displaystyle (c)\) B walks faster than A.(d) Who gets home first — compare where the lines end along the time axis. This is a different comparison from (a): (a) asked about the height the lines end at (distance from school), this asks about the time value at which each line ends (arrival clock time). In Fig. $\displaystyle 2.9$ both lines terminate at the same value of \(\displaystyle t \). \(\displaystyle (d)\) A and B reach home at the same time.(e) Who overtakes whom, and how many times — find where the two lines cross. Two children are at the same place on the road at the same instant exactly where their two \(\displaystyle x \)-\(\displaystyle t \) lines intersect — that is the geometric meaning of "overtaking" on this kind of graph. A left school first, so for a while A is farther down the road than B (B hasn't even started yet). Because B walks faster, B's position eventually catches up to and passes A's position. The two lines in Fig. $\displaystyle 2.9$ cross exactly one time, at a single point between the school and the homes, and after that crossing B stays ahead of A for the rest of the walk (consistent with B, who is slower to leave but quicker on foot, still arriving no later than A, from part (d)). \(\displaystyle (e)\) B overtakes A on the road once.Answer: (a) A lives closer to school than B; (b) A starts from school earlier than B; (c) B walks faster than A; (d) A and B reach home at the same time; (e) B overtakes A on the road once.

An x-t graph is built segment by segment: work out how long each leg of the journey takes from \(\displaystyle \text{speed} = \dfrac{\text{distance}}{\text{time}} \), then join the points. There are three legs here — walking out, standing still at the office, and riding back — and each one is a straight line on the graph because the speed is constant within it.Take \(\displaystyle t = 0 \) at $\displaystyle 9$:$\displaystyle 00$ am (when she leaves home) and let \(\displaystyle x \) be her position measured from home, in km, along the road toward the office.Leg $\displaystyle 1$ — walking to the office. Using \(\displaystyle \text{time} = \dfrac{\text{distance}}{\text{speed}} \), with distance \(\displaystyle = 2.5 \text{ km} \) and speed \(\displaystyle = 5 \text{ km h}^{-1} \): \[t_1 = \frac{2.5 \text{ km}}{5 \text{ km h}^{-1}} = 0.5 \text{ h} = 30 \text{ min} \] So she reaches the office at $\displaystyle 9$:$\displaystyle 30$ am, at \(\displaystyle x = 2.5 \) km. This leg is the straight line from \(\displaystyle (0,\,0) \) to \(\displaystyle (0.5\text{ h},\,2.5\text{ km}) \), with slope \[\text{slope} = \frac{2.5 - 0}{0.5 - 0} = +5 \text{ km h}^{-1}, \] which is just her walking velocity (positive, since she is moving away from home) — the slope of an x-t graph is always the velocity, not a separate quantity to compute.Leg $\displaystyle 2$ — at the office. She stays from $\displaystyle 9$:$\displaystyle 30$ am to $\displaystyle 5$:$\displaystyle 00$ pm, a duration of \[t_2 = 8\text{h}\,00 - 0\text{h}\,30 = 7.5 \text{ h}. \] Her position does not change, so this leg is a horizontal line at \(\displaystyle x = 2.5 \) km, from \(\displaystyle (0.5\text{ h}, 2.5\text{ km}) \) to \(\displaystyle (8.0\text{ h}, 2.5\text{ km}) \). A flat line on an x-t graph means zero velocity — being at rest, not "no motion to plot."Leg $\displaystyle 3$ — the auto ride home. Same distance, much higher speed: \[t_3 = \frac{2.5 \text{ km}}{25 \text{ km h}^{-1}} = 0.1 \text{ h} = 6 \text{ min} \] Leaving the office at $\displaystyle 5$:$\displaystyle 00$ pm (\(\displaystyle t = 8.0 \) h), she is home by $\displaystyle 5$:$\displaystyle 06$ pm (\(\displaystyle t = 8.1 \) h), at \(\displaystyle x = 0 \). This leg runs from \(\displaystyle (8.0\text{ h}, 2.5\text{ km}) \) to \(\displaystyle (8.1\text{ h}, 0\text{ km}) \), with slope \[\text{slope} = \frac{0 - 2.5}{8.1 - 8.0} = -25 \text{ km h}^{-1}. \] The sign is negative because she is now moving back toward home, and the aside worth noting: this segment must be drawn five times as steep as Leg $\displaystyle 1$ (steeper by the same factor, $\displaystyle 25$/$\displaystyle 5$ = $\displaystyle 5$, that the auto's speed exceeds the walking speed), even though it covers the same $\displaystyle 2.5$ km — steepness on an x-t graph tracks speed, and people often draw the return line only "somewhat" steeper instead of proportionally steeper.Putting it on paper — the four corner points of the graph: Joining them in order gives a gently rising line, then a long flat line, then a very steep falling line back to the axis.Suitable scales: since the total elapsed time is $\displaystyle 8.1$ h and the maximum displacement is $\displaystyle 2.5$ km, a convenient choice is With these scales Leg $\displaystyle 1$ rises gently over half a square, Leg $\displaystyle 2$ is a long flat stretch across roughly $\displaystyle 7.5$ squares, and Leg $\displaystyle 3$ drops almost vertically within a tenth of a square — visually showing that the return journey, though covering the same distance, takes a small fraction of the time.Answer: A three-segment x-t graph — straight line from ($\displaystyle 9$:$\displaystyle 00$ am, $\displaystyle 0$ km) to ($\displaystyle 9$:$\displaystyle 30$ am, $\displaystyle 2.5$ km) with slope +$\displaystyle 5$ km h⁻¹; horizontal line from ($\displaystyle 9$:$\displaystyle 30$ am, $\displaystyle 2.5$ km) to ($\displaystyle 5$:$\displaystyle 00$ pm, $\displaystyle 2.5$ km); then a much steeper straight line from ($\displaystyle 5$:$\displaystyle 00$ pm, $\displaystyle 2.5$ km) down to ($\displaystyle 5$:$\displaystyle 06$ pm, $\displaystyle 0$ km) with slope −$\displaystyle 25$ km h⁻¹ (five times as steep as the outward leg, since the auto is five times as fast).

The catch is that the drunkard can fall into the pit in the middle of a forward run — you cannot just use his average speed over a full cycle, because that would let him "borrow" distance he hasn't actually covered yet.Setting up one cycle. Every cycle of his walk is: $\displaystyle 5$ steps forward ($\displaystyle 5$ m in $\displaystyle 5$ s), then $\displaystyle 3$ steps backward ($\displaystyle 3$ m in $\displaystyle 3$ s) — one step of $\displaystyle 1$ m taking $\displaystyle 1$ s. So one full cycle takes \[T_{\text{cycle}} = 5\,\text{s} + 3\,\text{s} = 8\,\text{s}, \] and the net displacement after one full cycle is \[\Delta x = (+5\,\text{m}) + (-3\,\text{m}) = +2\,\text{m}. \] This +$\displaystyle 2$ m every $\displaystyle 8$ s is the drift of the walk — but it is not the speed at which he approaches the pit, because all of that $\displaystyle 2$ m of progress happens during the forward $\displaystyle 5$ s, and then part of it is given back during the backward $\displaystyle 3$ s.Tracking the position after each full cycle. Since each complete cycle adds $\displaystyle 2$ m to where the previous cycle left off, the position at the end of \(\displaystyle n\) complete cycles (i.e., at time \(\displaystyle t = 8n\) s) is \[x(8n) = 2n \ \text{m}. \] So: The pit is met during a forward run, not at the end of a cycle. In the forward part of cycle \(\displaystyle n+1\), starting from \(\displaystyle x = 2n\), he moves $\displaystyle 1$ m forward every second, so during that stretch \[x(t) = 2n + \big(t - 8n\big), \qquad 8n \le t \le 8n+5. \] The farthest he gets in that forward stretch, at \(\displaystyle t = 8n+5\), is \(\displaystyle x = 2n+5\). We need the smallest \(\displaystyle n\) for which this reaches the pit at $\displaystyle 13$ m: \[2n + 5 \ge 13 \ \Rightarrow\ n \ge 4. \] Taking \(\displaystyle n = 4\): at \(\displaystyle t = 32\) s he is at \(\displaystyle x = 8\) m (start of the 5th forward run). He then needs \[13\,\text{m} - 8\,\text{m} = 5\,\text{m} \] more, walked forward at $\displaystyle 1$ m/s, which takes exactly $\displaystyle 5$ s more. So he reaches \(\displaystyle x=13\) m — the pit — right at the end of that forward run, having taken all $\displaystyle 5$ of its steps, with no backward steps left to pull him back: \[t = 8(4) + 5 = 37\,\text{s}. \]Why the "average speed" shortcut fails. If you divide $\displaystyle 13$ m by the average drift of \(\displaystyle 2\,\text{m}/8\,\text{s} = 0.25\,\text{m/s}\), you get \(\displaystyle 13/0.25 = 52\) s — far too long. That number describes where he'd be if the forward-and-back oscillation were smoothed out, but the pit doesn't care about his average trend; it only cares about the instant his actual foot first lands at $\displaystyle 13$ m, which happens during a forward run, well before the smoothed-out estimate.The x-t graph. Plotting position against time gives a saw-tooth: a steep rising segment of slope \(\displaystyle +1\,\text{m/s}\) for $\displaystyle 5$ s, then a gentler falling segment of slope \(\displaystyle -1\,\text{m/s}\) for $\displaystyle 3$ s, repeating with the whole pattern drifting upward by $\displaystyle 2$ m each cycle. The turning points (peaks and troughs) are: \[(0,0) \to (5,5) \to (8,2) \to (13,7) \to (16,4) \to (21,9) \to (24,6) \to (29,11) \to (32,8) \to (37,13). \] Drawing a horizontal dashed line at \(\displaystyle x = 13\) m shows it is first touched exactly at the tip of the rising segment ending at \(\displaystyle t = 37\) s — the graph never crosses \(\displaystyle x=13\) m earlier, and this last rising segment ends exactly on the line rather than overshooting it, confirming both the graphical and the algebraic result agree.Answer: The drunkard falls into the pit at \(\displaystyle t = 37\) s (after $\displaystyle 4$ complete forward-backward cycles plus $\displaystyle 5$ more forward steps).

Convert everything to one consistent set of units first — kilometres per hour will not mix with metres.The car's speed is given in km h⁻¹, but the stopping distance is in metres, so convert the speed to m s⁻¹ before using any equation of motion.\[u = 126 \text{ km h}^{-1} = 126 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 35 \text{ m s}^{-1} \]The car comes to rest, so the final velocity is \(\displaystyle v = 0 \), and it travels a distance \(\displaystyle s = 200 \text{ m} \) while doing so.Step $\displaystyle 1$: Find the retardation using \(\displaystyle v^2 = u^2 + 2as\).This is one of the three equations of motion for uniform acceleration, where \(\displaystyle a\) is the (constant) acceleration — here it will come out negative, which is exactly what "retardation" means: acceleration directed opposite to the velocity.\[v^2 = u^2 + 2as \]Substitute \(\displaystyle v = 0\), \(\displaystyle u = 35 \text{ m s}^{-1}\), \(\displaystyle s = 200 \text{ m}\):\[0 = (35)^2 + 2a(200) \]\[0 = 1225 + 400a \]\[a = -\frac{1225}{400} = -3.0625 \text{ m s}^{-2} \]The negative sign shows the acceleration acts opposite to the direction of motion — that is the retardation. Rounding to three significant figures (matching the three figures in $\displaystyle 126$ km h⁻¹):\[\text{Retardation} = 3.06 \text{ m s}^{-2} \]A common slip here is to plug in \(\displaystyle u\) still in km h⁻¹ — the units on both sides of an equation of motion must match, so the conversion to m s⁻¹ has to happen before this step, not after.Step $\displaystyle 2$: Find the time using \(\displaystyle v = u + at\).\[v = u + at \]\[0 = 35 + (-3.0625)t \]\[t = \frac{35}{3.0625} = 11.428\ldots \text{ s} \]Rounding to three significant figures:\[t \approx 11.4 \text{ s} \]Answer: The retardation is $\displaystyle 3.06$ m s⁻² (i.e., acceleration of magnitude $\displaystyle 3.06$ m s⁻² directed opposite to the car's motion), and the car takes about $\displaystyle 11.4$ s to stop.

During the whole flight — rising, at the top, and falling — the only force on the ball is its own weight, so its acceleration is g, directed vertically downward, at every instant.(a) Direction of acceleration during the upward motionOnce the ball leaves the player's hand, the only force acting on it is gravity, \(\displaystyle mg \), pulling it toward the earth. By Newton's second law, \(\displaystyle a = F/m \), the acceleration is \(\displaystyle g \), directed vertically downward — the same direction throughout the flight. The acceleration does not reverse or vanish just because the ball happens to be moving upward; velocity and acceleration are independent quantities here.(b) Velocity and acceleration at the highest pointAt the highest point the ball has stopped rising and is about to start falling, so its velocity is momentarily zero: \[v = 0 \]The acceleration is not zero, though. Weight acts on the ball whether it is moving or momentarily at rest, so \[a = g = 9.8 \ \text{m s}^{-2}, \ \text{directed vertically downward.} \]This is the step people get wrong: zero velocity does not mean zero acceleration. The ball is instantaneously at rest but is still being pulled downward at \(\displaystyle 9.8 \ \text{m s}^{-2} \) — that is exactly why it does not stay at the top but starts to fall.(c) Signs of position, velocity and accelerationTake \(\displaystyle x = 0 \) and \(\displaystyle t = 0 \) at the highest point, with the positive \(\displaystyle x \)-axis pointing vertically downward.Upward motion (before \(\displaystyle t = 0 \), ball rising from the hand toward the top): At the highest point (\(\displaystyle t = 0 \)): \(\displaystyle x = 0 \), \(\displaystyle v = 0 \), but \(\displaystyle a = +g \), still positive. Acceleration never becomes zero at any point in the flight.Downward motion (after \(\displaystyle t = 0 \), ball falling from the top back to the hand): So position is positive through nearly the whole flight (zero only for the instant at the top), acceleration is positive and constant (\(\displaystyle +g \)) throughout, and it is only the velocity that changes sign — negative going up, zero at the top, positive coming down.(d) Maximum height and total time of flightHeight. Use \(\displaystyle v^2 = u^2 - 2gh \), where \(\displaystyle u = 29.4 \ \text{m s}^{-1} \) is the launch speed, \(\displaystyle v = 0 \) is the speed at the highest point, and \(\displaystyle h \) is the height risen (taking upward as positive for this scalar equation): \[0 = u^2 - 2gh \] \[h = \frac{u^2}{2g} = \frac{(29.4 \ \text{m s}^{-1})^2}{2 \times 9.8 \ \text{m s}^{-2}} = \frac{864.36 \ \text{m}^2\text{s}^{-2}}{19.6 \ \text{m s}^{-2}} = 44.1122\ldots \ \text{m} \]Rounding once, at the end, to three significant figures (matching the three figures given in \(\displaystyle 29.4 \ \text{m s}^{-1} \)): \[h = 44.1 \ \text{m} \]Time. Use \(\displaystyle v = u - gt \) with \(\displaystyle v = 0 \) at the top, to get the time to rise: \[0 = u - g\,t_{\text{up}} \] \[t_{\text{up}} = \frac{u}{g} = \frac{29.4 \ \text{m s}^{-1}}{9.8 \ \text{m s}^{-2}} = 3.0 \ \text{s} \]The ball retraces the same path under the same constant acceleration, so the fall takes exactly as long as the rise. The total time before the ball is back in the player's hands is \[T = 2\,t_{\text{up}} = 2 \times 3.0 \ \text{s} = 6.0 \ \text{s} \]Answer: (a) Vertically downward throughout the upward motion. (b) Velocity = $\displaystyle 0$; acceleration = $\displaystyle 9.8$ m s⁻², directed vertically downward. (c) With x = $\displaystyle 0$, t = $\displaystyle 0$ at the highest point and downward taken as positive: position is positive during both the upward and downward journeys (zero only at the highest point); velocity is negative while rising, zero at the top, and positive while falling; acceleration is positive (= +g) throughout and never zero. (d) Maximum height = $\displaystyle 44.1$ m; total time to return to the player's hands = $\displaystyle 6.0$ s.

In one dimension there is no "sideways": a velocity can only change its size or flip its sign — and it cannot flip while the speed stays constant. That one fact decides part (c); the other three follow straight from the definitions.The definitions used throughout, stated once: (a) A particle with zero speed at an instant may have non-zero acceleration at that instant — TRUE.Acceleration measures how fast \(\displaystyle v\) is changing, not how big \(\displaystyle v\) is. A velocity that is momentarily zero can still be changing rapidly.Example: a stone thrown vertically upward at \(\displaystyle u = 20\ \mathrm{m\,s^{-1}}\), taking upward as positive, so \(\displaystyle a=-g=-9.8\ \mathrm{m\,s^{-1}}\) throughout the flight. Using \(\displaystyle v=u+at\),\[0 = 20\ \mathrm{m\,s^{-1}} + \left(-9.8\ \mathrm{m\,s^{-2}}\right)t \quad\Rightarrow\quad t = \frac{20\ \mathrm{m\,s^{-1}}}{9.8\ \mathrm{m\,s^{-2}}} = 2.0408\ldots\ \mathrm{s} \approx 2.0\ \mathrm{s} \](two significant figures, set by the \(\displaystyle 20\ \mathrm{m\,s^{-1}}\) of the data). At that instant the speed is exactly \(\displaystyle 0\ \mathrm{m\,s^{-1}}\), yet the acceleration is still \(\displaystyle 9.8\ \mathrm{m\,s^{-2}}\) directed downward — gravity does not switch off at the top of the flight. If it did, the stone would hang there.(b) A particle with zero speed may have non-zero velocity — FALSE.Speed is defined as \(\displaystyle |v|\). So \(\displaystyle |v| = 0\) forces \(\displaystyle v = 0\): the only number whose magnitude is zero is zero itself. There is nothing physical to argue about here; it is the definition.The converse is what confuses people, and the converse is the true statement: a non-zero velocity always has non-zero speed, and a zero velocity always has zero speed. The two vanish together.(c) A particle with constant speed must have zero acceleration — TRUE.Let the constant speed be \(\displaystyle |v| = c\). In one dimension that leaves only two possible velocities at any instant, \(\displaystyle v=+c\) or \(\displaystyle v=-c\). So the velocity is either constant, or it jumps between \(\displaystyle +c\) and \(\displaystyle -c\).If it is constant, \(\displaystyle a=\dfrac{dv}{dt}=0\), and the statement holds.Could it jump instead? Suppose the particle is moving with \(\displaystyle v=+c\) and reverses to \(\displaystyle v=-c\) in a time interval \(\displaystyle \Delta t\). The average acceleration over that interval is\[\bar{a}=\frac{\Delta v}{\Delta t}=\frac{(-c)-(+c)}{\Delta t}=\frac{-2c}{\Delta t} \]For the speed to have stayed at \(\displaystyle c\) the whole time, the reversal must be instantaneous — the velocity is never allowed to take any intermediate value, including the value \(\displaystyle 0\) it would have to pass through. Setting \(\displaystyle \Delta t \to 0\) with \(\displaystyle \Delta v = -2c\) fixed gives \(\displaystyle |\bar{a}| \to \infty\). An infinite acceleration needs an infinite force, so no real particle does this. A ball bouncing off a wall looks like the exception, but it is not: during the contact the ball genuinely slows to zero and speeds up the other way, so its speed is not constant while it turns around.Every physically realisable $\displaystyle 1$-D motion at constant speed therefore has constant velocity, and hence \(\displaystyle a=0\).The step people get wrong here: the usual counterexample offered against (c) is uniform circular motion, where the speed is constant but the acceleration is \(\displaystyle v^2/r\) toward the centre. That is a genuine counterexample — in two dimensions. The stem restricts the particle to one-dimensional motion, and in $\displaystyle 1$-D there is no direction available for the velocity to turn into. Reaching for the circular-motion example here is importing a freedom the problem has taken away.(d) A particle with a positive value of acceleration must be speeding up — FALSE.Speeding up means \(\displaystyle |v|\) is increasing, and that happens only when \(\displaystyle v\) and \(\displaystyle a\) have the same sign. A positive acceleration acting on a negative velocity makes the particle slow down.Example: throw a stone upward at \(\displaystyle 20\ \mathrm{m\,s^{-1}}\), but this time choose downward as the positive direction. Then during the upward flight \(\displaystyle v = -20\ \mathrm{m\,s^{-1}}\) (negative, because it moves against the chosen positive direction) while \(\displaystyle a = +9.8\ \mathrm{m\,s^{-2}}\) (positive). After \(\displaystyle 1.0\ \mathrm{s}\),\[v = -20\ \mathrm{m\,s^{-1}} + \left(+9.8\ \mathrm{m\,s^{-2}}\right)\left(1.0\ \mathrm{s}\right) = -10.2\ \mathrm{m\,s^{-1}} \approx -10\ \mathrm{m\,s^{-1}} \](two significant figures). The speed has fallen from \(\displaystyle 20\ \mathrm{m\,s^{-1}}\) to about \(\displaystyle 10\ \mathrm{m\,s^{-1}}\) even though the acceleration is positive. The particle is slowing down.So (d) is true only in the special case where the positive direction has been chosen along the direction of motion, i.e. when \(\displaystyle v>0\). Nothing in the statement guarantees that, and the sign of \(\displaystyle a\) alone is a fact about your choice of axis, not about the motion.Answer: (a) True; (b) False; (c) True; (d) False.

A speed–time graph never dips below the time-axis — even though the ball reverses direction at the floor and again at the top of each bounce, the curve is built from straight segments of slope \(\displaystyle \pm g\), joined by sudden vertical drops where a collision instantly cuts the speed.Step $\displaystyle 1$ — speed just before the first impact. The ball falls from rest, so use \(\displaystyle v^2 = u^2 + 2gh \) (third equation of motion) with \(\displaystyle u = 0\), \(\displaystyle h = 90\ \text{m}\), and \(\displaystyle g = 9.8\ \text{m s}^{-2}\): \[v_1 = \sqrt{2gh} = \sqrt{2(9.8\ \text{m s}^{-2})(90\ \text{m})} = \sqrt{1764\ \text{m}^2\text{s}^{-2}} = 42.0\ \text{m s}^{-1} \]Since it starts from rest, \(\displaystyle v = gt\), so the time to reach the floor is \[t_1 = \frac{v_1}{g} = \frac{42.0\ \text{m s}^{-1}}{9.8\ \text{m s}^{-2}} = 4.29\ \text{s} \]Segment $\displaystyle 1$ ($\displaystyle 0$ s to $\displaystyle 4.29$ s): the straight line \(\displaystyle v = (9.8\ \text{m s}^{-2})\,t\), rising from the origin \(\displaystyle (0,0)\) to \(\displaystyle (4.29\ \text{s},\ 42.0\ \text{m/s})\). Slope \(\displaystyle +g\): the ball speeds up as it falls.Step $\displaystyle 2$ — the collision. The ball doesn't stop at the floor — it bounces back up, but "loses one-tenth of its speed" means the rebound speed is nine-tenths of the impact speed, not its mirror image: \[v_1' = \frac{9}{10}v_1 = 0.9 \times 42.0\ \text{m s}^{-1} = 37.8\ \text{m s}^{-1} \]On the graph this is a vertical drop at \(\displaystyle t_1 = 4.29\ \text{s}\), straight down from $\displaystyle 42.0$ m/s to $\displaystyle 37.8$ m/s. The collision is treated as instantaneous, so the speed changes by a finite jump at a single instant of time — this is the one feature that isn't a straight line.Step $\displaystyle 3$ — rising after the bounce. Moving upward now, gravity decelerates the ball at the same rate \(\displaystyle g\), so its speed falls from $\displaystyle 37.8$ m/s to zero at the top of the bounce: \[t_{\text{top}} = t_1 + \frac{v_1'}{g} = 4.29\ \text{s} + \frac{37.8\ \text{m s}^{-1}}{9.8\ \text{m s}^{-2}} = 4.29\ \text{s} + 3.86\ \text{s} = 8.14\ \text{s} \]Segment $\displaystyle 2$ ($\displaystyle 4.29$ s to $\displaystyle 8.14$ s): straight line falling from \(\displaystyle (4.29\ \text{s},\ 37.8\ \text{m/s})\) to \(\displaystyle (8.14\ \text{s},\ 0)\), slope \(\displaystyle -g\).(Aside — this is the step people draw wrong: the ball's velocity has reversed sign here, but a speed–time graph plots \(\displaystyle |v|\), so this segment still sits above the axis, sloping down to zero — it is never drawn as a negative-going line.)Step $\displaystyle 4$ — falling back down to the second impact. From the top the ball falls again from rest, and by symmetry of projectile motion under gravity alone, it takes exactly as long to fall back to floor height as it took to rise: \[t_2 = t_{\text{top}} + \frac{v_1'}{g} = 8.14\ \text{s} + 3.86\ \text{s} = 12.0\ \text{s} \]Segment $\displaystyle 3$ ($\displaystyle 8.14$ s to $\displaystyle 12.0$ s): straight line rising from \(\displaystyle (8.14\ \text{s}, 0)\) to \(\displaystyle (12.0\ \text{s},\ 37.8\ \text{m/s})\), slope \(\displaystyle +g\).Notice \(\displaystyle t_2\) comes out to exactly \(\displaystyle 12.0\ \text{s}\) — the right edge of the window the question asks for lands precisely on the instant of the second collision with the floor. That is not a coincidence to fix; it is where the graph asked for should end.The full picture, t = $\displaystyle 0$ s to t = $\displaystyle 12$ s: So the trace looks like a tall rising line, a sharp step down, then one symmetric "V" that just reaches its next peak exactly as the clock hits $\displaystyle 12$ s. (Each later bounce would repeat this same V-shape, only $\displaystyle 9$/$\displaystyle 10$ as tall as the one before it, since every rebound speed is $\displaystyle 0.9$ times the previous one — but those bounces start only after \(\displaystyle t = 12\ \text{s}\) and lie outside what is asked for here.)Answer: three straight-line segments of slope \(\displaystyle \pm 9.8\ \text{m s}^{-2}\), joined by one vertical drop — rising from \(\displaystyle (0,0)\) to \(\displaystyle (4.29\ \text{s}, 42.0\ \text{m/s})\); dropping instantaneously to \(\displaystyle (4.29\ \text{s}, 37.8\ \text{m/s})\) at the 1st collision; falling to \(\displaystyle (8.14\ \text{s}, 0)\) at the top of the bounce; then rising back up to \(\displaystyle (12.0\ \text{s}, 37.8\ \text{m/s})\), which is exactly the instant of the 2nd collision with the floor.

Distance is how far you actually walked; displacement is how far you ended up from where you started. For motion along a straight line these are usually different numbers, and the question asks you to show why one is never smaller than the other.(a) Magnitude of displacement vs. total path lengthDisplacement is the vector \(\displaystyle \Delta x = x_{\text{final}} - x_{\text{initial}} \) — it depends only on the start point and the end point, not on what happened in between. Its magnitude is \(\displaystyle |\Delta x| \).Path length (also called distance) is the sum of every bit of length actually covered, forward and backward, over the whole interval. It is a scalar and it is always positive.Example (one dimension, along the x-axis): A particle starts at \(\displaystyle x = 0 \), moves to \(\displaystyle x = +10\ \text{m} \), then comes back to \(\displaystyle x = +4\ \text{m} \). \[\text{path length} = 10\ \text{m} + 6\ \text{m} = 16\ \text{m}. \]Here \(\displaystyle 16\ \text{m} > 4\ \text{m} \): the path length is much larger than the magnitude of the displacement, because the return trip added to the distance walked but subtracted from the net displacement.Why path length ≥ |displacement| in general: every step the particle takes, forward or backward, adds a positive amount to the path length. But steps in the backward direction cancel part of the forward displacement instead of adding to it. So the path length keeps accumulating while the net displacement can only grow, shrink, or reverse — it can never accumulate faster than the path length does. This is the common mistake: people call the two "distance" interchangeably, but distance never subtracts, while displacement does.Equality sign in (a): the two become equal only when the particle never reverses direction — it moves monotonically from \(\displaystyle x_i \) to \(\displaystyle x_f \) without ever doubling back. Then every bit of path length is also a bit of net displacement, so \[\text{path length} = |\Delta x|. \] In the example above, if the particle had gone straight from \(\displaystyle x=0 \) to \(\displaystyle x=4\ \text{m} \) without overshooting to \(\displaystyle 10\ \text{m} \) first, path length and \(\displaystyle |\Delta x| \) would both equal \(\displaystyle 4\ \text{m} \).(b) Magnitude of average velocity vs. average speedAverage velocity is defined using the displacement: \[\vec{v}_{\text{av}} = \frac{\Delta x}{\Delta t}, \] where \(\displaystyle \Delta x \) is the displacement over the time interval \(\displaystyle \Delta t \). Its magnitude is \(\displaystyle |\Delta x|/\Delta t \).Average speed is defined using the path length (this is the definition given in the question itself): \[v_{\text{av,speed}} = \frac{\text{total path length}}{\Delta t}. \]Since \(\displaystyle \Delta t \) is the same positive number in both formulas, and part (a) already established \[\text{path length} \geq |\Delta x|, \] dividing both sides by the same \(\displaystyle \Delta t > 0 \) preserves the inequality: \[\frac{\text{path length}}{\Delta t} \geq \frac{|\Delta x|}{\Delta t} \quad\Longrightarrow\quad \text{average speed} \geq |\vec{v}_{\text{av}}|. \]Same example, with a time interval attached: suppose the outward \(\displaystyle 10\ \text{m} \) took \(\displaystyle 2\ \text{s} \) and the return \(\displaystyle 6\ \text{m} \) took another \(\displaystyle 2\ \text{s} \), so \(\displaystyle \Delta t = 4\ \text{s} \).\[|\vec{v}_{\text{av}}| = \frac{|\Delta x|}{\Delta t} = \frac{4\ \text{m}}{4\ \text{s}} = 1\ \text{m/s}, \qquad v_{\text{av,speed}} = \frac{16\ \text{m}}{4\ \text{s}} = 4\ \text{m/s}. \]The average speed \(\displaystyle (4\ \text{m/s}) \) is greater than the magnitude of the average velocity \(\displaystyle (1\ \text{m/s}) \) — exactly as it must be, since it is built from the larger path length divided by the same time.This is the step people get wrong: they assume "speed" and "velocity" are just two names for the same number. They coincide only under a specific condition, given next.Equality sign in (b): average speed equals the magnitude of average velocity precisely when path length equals \(\displaystyle |\Delta x| \) — that is, when the motion is in one direction only throughout \(\displaystyle \Delta t \), with no reversal. If the same particle had gone straight from \(\displaystyle x=0 \) to \(\displaystyle x=4\ \text{m} \) in \(\displaystyle 4\ \text{s} \) without the detour to \(\displaystyle 10\ \text{m} \), both quantities would equal \(\displaystyle \dfrac{4\ \text{m}}{4\ \text{s}} = 1\ \text{m/s} \).Answer: In both cases the second quantity (path length in (a), average speed in (b)) is greater than or equal to the first, because path length accumulates every bit of motion while displacement only records the net change in position, and dividing by the same time interval carries this inequality over to average speed vs. the magnitude of average velocity. Equality holds in both (a) and (b) under the same condition: the particle moves in one direction only, without reversing, over the interval considered.

Average velocity uses displacement (start to end, in a straight line); average speed uses the total path length walked — they need not agree, and here they spectacularly don't.Take the direction from home to the market as positive.Step $\displaystyle 1$: Find the two leg-times.Speed is distance divided by time, so time is distance divided by speed.Going to market (distance \(\displaystyle d = 2.5\ \mathrm{km} \), speed \(\displaystyle v_1 = 5\ \mathrm{km\,h^{-1}} \)): \[t_1 = \frac{d}{v_1} = \frac{2.5\ \mathrm{km}}{5\ \mathrm{km\,h^{-1}}} = 0.5\ \mathrm{h} = 30\ \mathrm{min} \]Returning home (same distance, speed \(\displaystyle v_2 = 7.5\ \mathrm{km\,h^{-1}} \)): \[t_2 = \frac{d}{v_2} = \frac{2.5\ \mathrm{km}}{7.5\ \mathrm{km\,h^{-1}}} = \frac{1}{3}\ \mathrm{h} = 20\ \mathrm{min} \]So the whole round trip takes \(\displaystyle t_1+t_2 = 50\ \mathrm{min} \). This tells you exactly what's happening in each of the three windows below: $\displaystyle 30$ min is "just reached the market," $\displaystyle 50$ min is "just got home," and $\displaystyle 40$ min is "$\displaystyle 10$ minutes into the walk back."(i) $\displaystyle 0$ to $\displaystyle 30$ minAt \(\displaystyle t = 30\ \mathrm{min} \) the man has just arrived at the market for the first time — he hasn't turned around yet, so every step in this window is in the same direction.Displacement = distance walked = \(\displaystyle +2.5\ \mathrm{km} \) (toward the market). Elapsed time = $\displaystyle 0.5$ h.\[\text{average velocity} = \frac{\text{displacement}}{\text{time}} = \frac{2.5\ \mathrm{km}}{0.5\ \mathrm{h}} = 5\ \mathrm{km\,h^{-1}} \] \[\text{average speed} = \frac{\text{path length}}{\text{time}} = \frac{2.5\ \mathrm{km}}{0.5\ \mathrm{h}} = 5\ \mathrm{km\,h^{-1}} \]With no reversal, path length equals the magnitude of displacement, so the two quantities coincide.(ii) $\displaystyle 0$ to $\displaystyle 50$ min$\displaystyle 50$ min is exactly \(\displaystyle t_1+t_2 \) — the man has just walked back in through his own front door. His net displacement from where he started is zero.\[\text{average velocity} = \frac{\text{displacement}}{\text{time}} = \frac{0\ \mathrm{km}}{(50/60)\ \mathrm{h}} = 0\ \mathrm{km\,h^{-1}} \]This is the trap the question is built around: zero average velocity does not mean he stood still — he walked $\displaystyle 5$ km. Average speed uses the path length, which keeps adding up on the way back instead of cancelling:\[\text{average speed} = \frac{\text{path length}}{\text{time}} = \frac{2.5\ \mathrm{km}+2.5\ \mathrm{km}}{50/60\ \mathrm{h}} = \frac{5\ \mathrm{km}}{5/6\ \mathrm{h}} = 6\ \mathrm{km\,h^{-1}} \](iii) $\displaystyle 0$ to $\displaystyle 40$ minNow the window ends inside the return leg. He reaches the market at \(\displaystyle t=30\ \mathrm{min} \) (position \(\displaystyle +2.5\ \mathrm{km} \)), then walks back toward home for the remaining \(\displaystyle 40-30 = 10\ \mathrm{min} = \tfrac{1}{6}\ \mathrm{h} \) at \(\displaystyle 7.5\ \mathrm{km\,h^{-1}} \):\[\text{distance walked back} = 7.5\ \mathrm{km\,h^{-1}} \times \frac{1}{6}\ \mathrm{h} = 1.25\ \mathrm{km} \]So at \(\displaystyle t = 40\ \mathrm{min} \) he is at position \(\displaystyle 2.5\ \mathrm{km} - 1.25\ \mathrm{km} = 1.25\ \mathrm{km} \) from home — still on the market side of the road.Displacement (position at $\displaystyle 40$ min minus position at $\displaystyle 0$ min): \[\Delta x = 1.25\ \mathrm{km} - 0 = +1.25\ \mathrm{km} \] (positive, so still in the home-to-market direction — he hasn't gotten back far enough to cross the starting point).Total elapsed time = $\displaystyle 40$ min = \(\displaystyle \tfrac{2}{3}\ \mathrm{h} \).\[\text{average velocity} = \frac{1.25\ \mathrm{km}}{2/3\ \mathrm{h}} = 1.875\ \mathrm{km\,h^{-1}} \approx 1.9\ \mathrm{km\,h^{-1}},\ \text{directed from home toward the market} \]Path length, by contrast, keeps both legs (there's no subtracting the way back — he actually walked all of it): \[\text{path length} = 2.5\ \mathrm{km} + 1.25\ \mathrm{km} = 3.75\ \mathrm{km} \] \[\text{average speed} = \frac{3.75\ \mathrm{km}}{2/3\ \mathrm{h}} = 5.625\ \mathrm{km\,h^{-1}} \approx 5.6\ \mathrm{km\,h^{-1}} \](Rounded to two significant figures at the end, matching the two-significant-figure data — $\displaystyle 2.5$ km, $\displaystyle 5$ km h⁻¹, $\displaystyle 7.5$ km h⁻¹ — given in the problem.)Answer: (i) average velocity = $\displaystyle 5$ km h⁻¹, average speed = $\displaystyle 5$ km h⁻¹. (ii) average velocity = $\displaystyle 0$ (he is back home), average speed = $\displaystyle 6$ km h⁻¹. (iii) average velocity ≈ $\displaystyle 1.9$ km h⁻¹ directed from home toward the market, average speed ≈ $\displaystyle 5.6$ km h⁻¹.