NCERT Solutions · Class 11 Physics Motion in a Straight Line

Over a very short time interval, distance travelled and the size of the displacement become the same thing — because any smooth path looks straight once you zoom in close enough.Start from the definitions.Average speed over a time interval \(\displaystyle \Delta t \) is \[\text{average speed} = \frac{\Delta s}{\Delta t} \] where \(\displaystyle \Delta s \) is the actual path length (distance) covered. Average velocity over the same interval is \[\vec{v}_{avg} = \frac{\Delta \vec{x}}{\Delta t} \] where \(\displaystyle \Delta \vec{x} \) is the displacement — the straight-line vector from the starting point to the ending point, regardless of what path was taken between them.For a finite \(\displaystyle \Delta t \), a particle can wander — curve, double back, loop — between its start and end points, so the path length \(\displaystyle \Delta s \) can be longer than the straight-line distance \(\displaystyle |\Delta \vec{x}| \). That is exactly why average speed and the magnitude of average velocity can differ (this is the point made in Exercises $\displaystyle 2.9$ and $\displaystyle 2.10$): a car going around a curved road, or a body returning to its starting point, covers real distance while its net displacement is smaller, or even zero.Now take the limit that defines the instantaneous quantities: \[v_{inst} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt}, \qquad \vec{v}_{inst} = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt} \]As \(\displaystyle \Delta t \to 0 \), the two endpoints of the interval get pulled closer and closer together on the path. Over that shrinking stretch, whatever curve the motion traces is being sampled on a smaller and smaller scale — and any smooth curve, viewed close enough, is indistinguishable from its own tangent line (this is the same idea as a curved graph looking locally straight when you zoom in far enough). So in this limit, the actual path length \(\displaystyle \Delta s \) between the two nearby points and the straight-line chord joining them, \(\displaystyle |\Delta \vec{x}| \), converge to the same value: \[\lim_{\Delta t \to 0} \Delta s = \lim_{\Delta t \to 0} |\Delta \vec{x}| \]Dividing both sides by \(\displaystyle \Delta t \) and taking the limit, \[\lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \lim_{\Delta t \to 0} \frac{|\Delta \vec{x}|}{\Delta t} \] \[\Rightarrow \quad v_{inst} = |\vec{v}_{inst}| \]The freedom to "curve away" from the straight-line displacement is what separated speed from velocity magnitude in the average case — and that freedom disappears once the interval shrinks to an instant, because there is no room left for the path to bend within an infinitesimally short stretch. That is the whole distinction: it is a statement about finite intervals, and it vanishes as the interval goes to zero.A useful aside: this is why, for a car going round a curved track, the average speed over a lap can exceed the magnitude of average velocity by a lot, yet at any single instant during that lap, the speedometer reading (instantaneous speed) is exactly equal to the magnitude of the instantaneous velocity vector — the two never disagree instant by instant, only when averaged over a stretch of curved path.**Answer: The instantaneous speed equals the magnitude of the instantaneous velocity because, as \(\displaystyle \Delta t \to 0 \), the actual path length \(\displaystyle \Delta s \) traversed and the magnitude of the displacement \(\displaystyle |\Delta \vec{x}| \) become equal (any smooth path is locally straight over an infinitesimally short interval), so \(\displaystyle v_{inst} = \lim_{\Delta t \to 0} \Delta s/\Delta t \) and \(\displaystyle |\vec{v}_{inst}| = \lim_{\Delta t \to 0} |\Delta \vec{x}|/\Delta t \) coincide — unlike the average quantities, which can differ over a finite interval because the path may curve away from the straight-line displacement.

The idea: a graph can only describe a real motion if (i) it passes the vertical‑line test — at one instant the particle has exactly one position and one velocity — and (ii) the quantity plotted obeys its own definition (speed and total path length can never be negative, and path length can never shrink). Test each of the four graphs against those two rules.The two rules, stated properlyRule $\displaystyle 1$ (single‑valuedness). Position is a function of time, \(\displaystyle x = x(t) \), and so is velocity, \(\displaystyle v = v(t) \). At one instant \(\displaystyle t \) a particle is at one place and moving at one rate. So any vertical line drawn on the graph — a line "\(\displaystyle t = \) constant" — must cut the curve at most once.Rule $\displaystyle 2$ (definitions). Speed is the magnitude of the velocity vector, \[v_{\text{speed}} = |\vec{v}| \ \ge \ 0 , \] and a magnitude is never negative. Total path length \(\displaystyle s \) up to time \(\displaystyle t \) is the distance actually travelled, \[s(t) = \int_{0}^{t} |\vec{v}|\,\mathrm{d}t' \qquad\Longrightarrow\qquad \frac{\mathrm{d}s}{\mathrm{d}t} = |\vec{v}| \ \ge \ 0 . \] Because the slope \(\displaystyle \mathrm{d}s/\mathrm{d}t \) can never be negative, \(\displaystyle s \) can stay flat (particle at rest) or rise, but it can never come back down. Distance already covered cannot be un‑covered.Graph (a) — \(\displaystyle x \) plotted against \(\displaystyle t \)Reading the picture: the curve is a closed figure‑of‑eight through the origin — one loop up in the region of positive \(\displaystyle t \), positive \(\displaystyle x \), a second loop down in the region of negative \(\displaystyle t \), negative \(\displaystyle x \), with arrowheads showing the particle running round each loop.Because it is a closed loop, a vertical line drawn anywhere through the middle of the upper loop — say at the \(\displaystyle t \) of the loop's widest part — cuts the curve twice. That says the particle is at two different values of \(\displaystyle x \) at the same instant \(\displaystyle t \). One particle cannot be in two places at once, so Rule $\displaystyle 1$ is broken.Graph (b) — \(\displaystyle v \) plotted against \(\displaystyle t \)Reading the picture: the curve is a complete circle, centred close to the origin and straddling the \(\displaystyle v \)-axis, with an arrowhead near the top showing the sense in which it is traced.Again a closed curve: a vertical line through the circle meets it at two points, an upper and a lower one. That would give the particle two different velocities — one positive, one negative — at the same instant \(\displaystyle t \). Rule $\displaystyle 1$ is broken.Graph (c) — Speed plotted against \(\displaystyle t \)Reading the picture: a smooth wave that starts high on the left, dips below the \(\displaystyle t \)-axis into a trough just to the left of the origin, rises to a crest at positive \(\displaystyle t \), and falls again. A good part of the curve lies underneath the horizontal axis.Every point below the \(\displaystyle t \)-axis is a negative speed. But by Rule $\displaystyle 2$, \(\displaystyle v_{\text{speed}} = |\vec{v}| \ge 0 \); a magnitude cannot be less than zero. (Velocity in one dimension may certainly be negative — that just means "moving in the \(\displaystyle -x \) direction" — but the vertical axis here is labelled Speed, not velocity.) So this graph is impossible.Graph (d) — Total path length plotted against \(\displaystyle t \)Reading the picture: starting at the origin the line rises straight to a peak, then comes straight back down to zero on the \(\displaystyle t \)-axis, then rises straight again — a triangular, saw‑tooth shape.On the falling side the slope is negative, i.e. \(\displaystyle \mathrm{d}s/\mathrm{d}t < 0 \), and the graph returns to \(\displaystyle s = 0 \) as if the particle had never moved. Rule $\displaystyle 2$ forbids this: \(\displaystyle \mathrm{d}s/\mathrm{d}t = |\vec{v}| \ge 0 \), so total path length is a non‑decreasing quantity. Even if the particle turns round and walks back to its starting point, its displacement returns to zero but its path length keeps growing. So this graph is impossible.Collecting the results

Graph	What it plots	Rule broken
(a)	\(\displaystyle x \) vs \(\displaystyle t \)	two positions at one instant
(b)	\(\displaystyle v \) vs \(\displaystyle t \)	two velocities at one instant
(c)	speed vs \(\displaystyle t \)	speed goes negative
(d)	path length vs \(\displaystyle t \)	path length decreases

All four fail, each for its own reason.Answer: None of the four graphs can represent one‑dimensional motion — (a) and (b) are closed curves, so a vertical line \(\displaystyle t = \) constant meets them twice, giving the particle two positions (a) or two velocities (b) at the same instant, which is impossible; (c) dips below the \(\displaystyle t \)-axis, i.e. negative speed, whereas speed \(\displaystyle = |\vec{v}| \ge 0 \) always; and (d) falls back to zero, i.e. decreasing total path length, whereas \(\displaystyle \mathrm{d}s/\mathrm{d}t = |\vec{v}| \ge 0 \) means path length can never decrease.

A position–time graph is not a snapshot of the path — it only tells you how far along a fixed line the particle is at each instant. Reading "straight line, then parabola" off an \(\displaystyle x\text{-}t\) graph and treating that as the shape traced out in space is the mistake this question is built around.The problem statement itself says this is one-dimensional motion. By definition, a particle in one-dimensional motion moves only along a single straight line (call it the \(\displaystyle x\)-axis) for its entire history — it has nowhere else to go. So whatever the graph looks like, the actual trajectory in space is a straight line for all \(\displaystyle t\), both before and after \(\displaystyle t=0\). A curve cannot suddenly become "a parabolic path" in real space unless the motion picks up a second spatial dimension, which the problem rules out.What the two pieces of the graph really say is about how \(\displaystyle x\) depends on \(\displaystyle t\), not about geometry in space: \[x(t) = \tfrac{1}{2}at^{2} \] has exactly this shape, and it is the textbook signature of motion starting from rest with constant acceleration: its slope \(\displaystyle v=\dfrac{dx}{dt}=at\) grows linearly from zero, and \(\displaystyle a=\dfrac{dv}{dt}\) is constant. So the parabola is not a bent path — it is what constant acceleration along a straight line looks like when you plot position against time.So the correct statement is: the particle is at rest at the origin for \(\displaystyle t<0\), and from \(\displaystyle t=0\) onward it moves away from the origin along the same straight line with continuously increasing speed, because a constant force has begun to act on it. It never leaves that line.A suitable physical context, then, is any situation where an object is held at rest and then released into uniformly accelerated motion along a straight track — for example, a block resting on a smooth horizontal surface up to \(\displaystyle t=0\), which is then pulled by a constant force (a string running over a pulley to a falling weight, say) that switches on at \(\displaystyle t=0\); or, equally, a ball held at rest at the top of a smooth straight incline and let go at \(\displaystyle t=0\), after which it slides down the incline under the constant component of gravity along the slope. In either case the path is a straight line throughout, the particle is at rest for \(\displaystyle t<0\), and its position grows as \(\displaystyle t^{2}\) for \(\displaystyle t>0\) — reproducing exactly the graph described, with no parabolic path anywhere in space.**Answer: No — an \(\displaystyle x\text{-}t\) graph shows how the position varies with time, not the shape of the path, and since the motion is stated to be one-dimensional, the particle actually moves along one and the same straight line for all \(\displaystyle t\). The graph only shows that the particle is at rest at the origin for \(\displaystyle t<0\) and then, from \(\displaystyle t=0\), moves along that line with uniformly increasing speed, i.e., \(\displaystyle x \propto t^{2}\), consistent with a constant force switching on at \(\displaystyle t=0\) — as when a body at rest is released to slide down a straight, smooth incline (or is pulled by a constant force along a straight track).

The speed that matters for the damage is the bullet's speed relative to the thief's car — not its speed relative to the ground, and not its muzzle speed relative to the van.Whatever hits the car does so with the relative velocity between the bullet and the car; that is what determines the impact. So the plan is: ($\displaystyle 1$) find the bullet's velocity relative to the ground, then ($\displaystyle 2$) subtract the thief's car's velocity to get the bullet's velocity relative to the car.Step $\displaystyle 1$ — write down what is given, in consistent units.Speed of police van (relative to ground): \(\displaystyle v_p = 30 \text{ km h}^{-1} \)Speed of thief's car (relative to ground): \(\displaystyle v_t = 192 \text{ km h}^{-1} \)Muzzle speed of bullet relative to the van: \(\displaystyle v_{bp} = 150 \text{ m s}^{-1} \)All three motions are along the same straight line (the highway), in the same direction, so ordinary algebraic addition/subtraction of velocities applies — no vector components are needed, only a sign convention (take the common direction of travel as positive).Convert the km h\(\displaystyle ^{-1}\) speeds to m s\(\displaystyle ^{-1}\) using \(\displaystyle 1 \text{ km h}^{-1} = \dfrac{5}{18} \text{ m s}^{-1} \):\[v_p = 30 \times \frac{5}{18} \text{ m s}^{-1} = \frac{25}{3} \text{ m s}^{-1} \]\[v_t = 192 \times \frac{5}{18} \text{ m s}^{-1} = \frac{160}{3} \text{ m s}^{-1} \]Step $\displaystyle 2$ — get the bullet's velocity relative to the ground.The muzzle speed \(\displaystyle v_{bp} = 150 \text{ m s}^{-1} \) is the bullet's velocity relative to the van that fires it, not relative to the road. Since the van itself is moving at \(\displaystyle v_p \) in the same direction, the bullet's velocity relative to the ground is the plain sum (this is the one-dimensional case of \(\displaystyle \vec{v}_{bullet,\, ground} = \vec{v}_{bullet,\, van} + \vec{v}_{van,\, ground} \), the same rule used for adding relative velocities along a line):\[v_b = v_{bp} + v_p = 150 + \frac{25}{3} = \frac{450+25}{3} = \frac{475}{3} \text{ m s}^{-1} \]Step $\displaystyle 3$ — get the bullet's velocity relative to the thief's car.The velocity of the bullet relative to the thief's car is\[v_{bt} = v_b - v_t \]This is the aside worth flagging: it is easy to stop at \(\displaystyle v_b \) (the bullet's ground speed) and call that "the answer," but a bullet's ground speed does not damage a car that is itself moving — only the relative speed between the two does, which is exactly what the question's note is pointing at.\[v_{bt} = \frac{475}{3} - \frac{160}{3} = \frac{315}{3} = 105 \text{ m s}^{-1} \]The result is a whole number, so no rounding decision is needed — all three given quantities ($\displaystyle 30$, $\displaystyle 192$, $\displaystyle 150$) combine to give exactly $\displaystyle 105$ m s\(\displaystyle ^{-1}\).Direction: \(\displaystyle v_{bt} \) came out positive (using the direction of travel of both vehicles as positive), so the bullet still approaches the thief's car from behind, along the direction the vehicles are moving — it has not been "left behind" by the faster car.Answer: The bullet hits the thief's car at a relative speed of $\displaystyle 105$ m s\(\displaystyle ^{-1}\), directed along the common direction of motion of the vehicles.

The idea: a graph is a story about slopes, and the corners are the events. On an \(\displaystyle x\)–\(\displaystyle t\) graph the slope is the velocity, \(\displaystyle v=\dfrac{dx}{dt}\); on a \(\displaystyle v\)–\(\displaystyle t\) graph the slope is the acceleration, \(\displaystyle a=\dfrac{dv}{dt}\). So a straight sloping piece means uniform velocity, a horizontal piece on an \(\displaystyle x\)–\(\displaystyle t\) graph means at rest, and a sharp corner or a vertical jump means the velocity changed almost instantaneously — in the real world that only happens when something is hit, i.e. a large force acting for a very short contact time. Read each graph for three things: the sign of the quantity, whether it is constant, and where the corners are.None of the three axes carry numbers, so below I quote everything in "figure units" (pixels of the printed graph) measured against that graph's own axes — you can check every one of them against the same picture.Graph (a) — \(\displaystyle x\) against \(\displaystyle t\): a ball kicked across a smooth floor between two walls.What I read off the picture: \[\frac{v_2}{v_1}=\frac{0.61}{0.71}\approx 0.85 \] i.e. it comes back about \(\displaystyle 15\%\) slower than it went. (Equivalently, on the picture the fall through the same height \(\displaystyle h\) takes about \(\displaystyle 1.2\) times as long as the climb.) Since every piece is straight, the acceleration is zero everywhere except at the three corners, where it is huge and brief. That is the signature of collisions.A situation that produces exactly this: a ball lying at rest on a smooth (frictionless) floor between two walls is given a kick. It travels at constant velocity \(\displaystyle +0.71\) units towards the wall at A, rebounds elastically-but-not-quite (leaving with \(\displaystyle 0.85\) of its speed, now directed along \(\displaystyle -x\)), coasts back past its starting point, and hits a second wall at \(\displaystyle x\approx-0.55\,h\) which stops it dead — the collision there is completely inelastic, so it remains at rest. The floor must be smooth, otherwise friction would bend the straight segments into curves.Graph (b) — \(\displaystyle v\) against \(\displaystyle t\): a ball bouncing on the floor.What I read off the picture: Check that this really is free flight between the jumps. For a projectile launched upward with speed \(\displaystyle u\), the time to return is\[T=\frac{2u}{g}\qquad\Rightarrow\qquad T=\frac{2(64)}{2.25}=57\ \text{units}, \]which is exactly the measured gap after the first jump; using \(\displaystyle u=26\) gives \(\displaystyle T=23\) units, again the measured value. The graph is internally consistent with gravity alone.So the situation is a ball dropped (or thrown up) and bouncing on the floor. Between bounces only gravity acts, so \(\displaystyle v\) falls linearly at \(\displaystyle -g\); each contact with the floor lasts so short a time on this scale that it shows as a vertical jump in which the velocity flips from downward to upward; energy is lost in each bounce, so the ball leaves with less speed than it arrived, the rises get lower, the flights get shorter, and the ball finally settles on the floor. Note that the first line crosses \(\displaystyle v=0\) exactly on the vertical axis, so the clock's zero has been put at the top of the first flight — the graph starts while the ball is still on its way up.Graph (c) — \(\displaystyle a\) against \(\displaystyle t\): a ball struck by a bat.What I read off the picture: Meaning: \(\displaystyle a=0\) means the velocity is constant, so the body coasts at uniform velocity, receives one short burst of large positive acceleration, and then coasts again. The change in velocity is the area under the bump,\[\Delta v=\int a\,dt \;>\;0, \]so the body ends up moving faster in the \(\displaystyle +x\) direction than it started.A situation that produces this: a ball moving with uniform velocity along a smooth horizontal surface is struck by a bat (equally, a rolling football given one kick). The bat is in contact for only a few milliseconds, and during that contact the acceleration is enormous and directed along \(\displaystyle +x\); before and after, nothing acts along the direction of motion, so \(\displaystyle a=0\). The bump is rounded rather than an infinitely thin spike because the contact, though brief, is not instantaneous — the force builds up and dies away as the ball squashes and springs back. The motion must be horizontal (or the graph is of the horizontal component), because a vertical motion would show a constant non-zero \(\displaystyle a=-g\) instead of the flat zero line.Answer: (a) A ball at rest on a smooth floor is kicked at the first corner, runs at constant velocity \(\displaystyle +0.71\) units to a wall at A, rebounds there with about \(\displaystyle 0.85\) of its speed now along \(\displaystyle -x\), travels back past its starting point, and is stopped dead by a second wall at about \(\displaystyle x=-0.55h\), where it stays at rest; (b) a ball bouncing on the floor — free flight under gravity alone (every segment parallel with slope \(\displaystyle \approx-2.3\) units \(\displaystyle =-g\), upward positive) interrupted by brief floor contacts that flip the velocity from \(\displaystyle -100\) to \(\displaystyle +64\), \(\displaystyle -64\) to \(\displaystyle +26\), \(\displaystyle -28\) to \(\displaystyle +16\), so the flights shorten (\(\displaystyle 57, 23, 17, 10, 7\) units) and the ball settles; (c) a ball coasting at uniform velocity along a smooth horizontal surface is struck by a bat — one short, large, positive acceleration pulse with \(\displaystyle a=0\) before and after, which raises its speed in the \(\displaystyle +x\) direction and leaves it coasting again.

In this x-t graph, everything you need is the shape of the curve at each instant: whether it sits above or below the axis (position), whether it is rising or falling (velocity), and the fact that a particle in SHM always accelerates back toward its mean position — never away from it.Three things to read off Fig. $\displaystyle 2.13$, one at a time.1. Position \(\displaystyle x\): the height of the curve above (positive) or below (negative) the \(\displaystyle t\)-axis. 2. Velocity \(\displaystyle v = dx/dt\): the slope of the curve — rising gives \(\displaystyle v>0\), falling gives \(\displaystyle v<0\). 3. Acceleration \(\displaystyle a\): in SHM the restoring acceleration always points back toward the equilibrium position \(\displaystyle x=0\), so it is opposite in sign to the displacement, \[a = -\omega^2 x . \] Wherever \(\displaystyle x>0\), \(\displaystyle a<0\); wherever \(\displaystyle x<0\), \(\displaystyle a>0\). Once you know the sign of \(\displaystyle x\) you already know the sign of \(\displaystyle a\) — you don't need to look at the curvature separately.Reading Fig. $\displaystyle 2.13$ itself: the curve crosses the \(\displaystyle t\)-axis moving downward at \(\displaystyle t=0\), reaches a minimum (trough) at \(\displaystyle t=0.5\ \text{s}\), climbs back through zero moving upward at \(\displaystyle t=1\ \text{s}\), reaches a maximum (crest) at \(\displaystyle t=1.5\ \text{s}\), and is back at zero at \(\displaystyle t=2\ \text{s}\) — one full oscillation every \(\displaystyle 2\ \text{s}\). The same pattern continues to the left of \(\displaystyle t=0\): there is a trough at \(\displaystyle t=-1.5\ \text{s}\) and an upward zero-crossing at \(\displaystyle t=-1\ \text{s}\).At \(\displaystyle t = 0.3\ \text{s}\): This lies between the downward zero-crossing at \(\displaystyle t=0\) and the trough at \(\displaystyle t=0.5\ \text{s}\) — the curve is below the axis and still falling toward that trough. At \(\displaystyle t = 1.2\ \text{s}\): This lies between the upward zero-crossing at \(\displaystyle t=1\ \text{s}\) and the crest at \(\displaystyle t=1.5\ \text{s}\) — the curve is above the axis and still climbing toward that crest. At \(\displaystyle t = -1.2\ \text{s}\): This lies between the trough at \(\displaystyle t=-1.5\ \text{s}\) and the upward zero-crossing at \(\displaystyle t=-1\ \text{s}\) — the curve is below the axis but already climbing back up out of the trough. Where people slip: at \(\displaystyle t=-1.2\ \text{s}\) it feels like \(\displaystyle x\) and \(\displaystyle v\) "ought" to match signs — they don't have to. Velocity is the slope at that instant, not the side of the axis the particle happens to be on. A particle can be on the negative side of the mean position while already moving back toward it, which is exactly the case here: \(\displaystyle x<0\) but \(\displaystyle v>0\). Acceleration, in contrast, only ever depends on which side of \(\displaystyle x=0\) the particle is on, never on which way it is moving — that is why \(\displaystyle a\) tracks \(\displaystyle x\) but \(\displaystyle v\) does not.Answer: at \(\displaystyle t=0.3\,\text{s}\): \(\displaystyle x<0,\ v<0,\ a>0\). At \(\displaystyle t=1.2\,\text{s}\): \(\displaystyle x>0,\ v>0,\ a<0\). At \(\displaystyle t=-1.2\,\text{s}\): \(\displaystyle x<0,\ v>0,\ a>0\).

On an x-t graph, average velocity over any interval is nothing but the slope of the straight line joining the curve's two end-points — and average speed is simply the size of that slope, not a separate quantity to hunt for.For an interval from \(\displaystyle t_1 \) to \(\displaystyle t_2 \),\[\bar{v} = \frac{x_2 - x_1}{t_2 - t_1} \]This is exactly the slope of the chord drawn between the points where the curve meets the interval's two boundaries. A steeply tilted chord (up or down) means a large change in position in that time, i.e. a large \(\displaystyle |\bar v|\); a nearly flat chord means the position hardly changed, i.e. a small \(\displaystyle |\bar v|\).A short aside, because the two words get mixed up constantly: velocity carries a sign (it tells you which way along the line the particle net-moved), while speed is always non-negative. Average speed equals \(\displaystyle |\bar v|\) only when the particle does not reverse direction inside that interval — if it did, the forward and backward parts of the motion would partly cancel in \(\displaystyle \Delta x\) while still adding up in the actual path length, making the true average speed larger than \(\displaystyle |\Delta x/\Delta t|\). Looking at Figure $\displaystyle 2.14$, the curve's peak (maximum \(\displaystyle x\)) occurs just after interval $\displaystyle 2$ ends, and its trough (minimum \(\displaystyle x\)) occurs just after interval $\displaystyle 3$ ends — so none of the three marked intervals contains a turning point, and the particle moves in one direction throughout each one. That means average speed does equal \(\displaystyle |\bar v|\) in all three intervals here, and the two questions (speed and sign of velocity) can be answered from the same three chords.Reading the steepness of each chord directly off the curve:Ranking the three chords by how tilted they are: interval $\displaystyle 3$ (steep fall) > interval $\displaystyle 1$ (steady climb) > interval $\displaystyle 2$ (almost flat, near the peak).Answer: Average speed is greatest in interval $\displaystyle 3$ and least in interval 2. The average velocity is positive in intervals $\displaystyle 1$ and $\displaystyle 2$, and negative in interval 3.

The whole question is answered by two features of a speed–time graph: the height of the curve is the speed, and the slope of the curve is the acceleration.Because the particle moves along one constant direction, and that direction is chosen positive, the speed and the velocity are the same number here (\(\displaystyle v \ge 0\) always). So Fig. $\displaystyle 2.15$ is also the velocity–time graph, and the two definitions we need are\[\bar a=\frac{v_2-v_1}{t_2-t_1}\qquad\text{and}\qquad a=\frac{dv}{dt}\]where \(\displaystyle \bar a\) is the average acceleration over an interval, \(\displaystyle v_1\) and \(\displaystyle v_2\) are the speeds at the start (\(\displaystyle t_1\)) and end (\(\displaystyle t_2\)) of that interval, and \(\displaystyle a\) is the instantaneous acceleration. Geometrically: \(\displaystyle \bar a\) is the slope of the straight chord joining the two ends of the interval, and \(\displaystyle a\) is the slope of the tangent at a single point.What is actually on the picture. The speed axis carries no numbers, so every reading has to be a comparison. I will use the height of the highest point on the graph, the peak \(\displaystyle D\), as the yardstick and call it \(\displaystyle v_D\); the time unit \(\displaystyle T\) is the spacing between the marks $\displaystyle 1$, $\displaystyle 2$ and 3. The three shaded strips are the three equal intervals: each is centred on a mark and each is about \(\displaystyle 0.12\,T\) wide. From the curve: Which interval has the greatest average acceleration in magnitude? Substituting each strip into \(\displaystyle \bar a=(v_2-v_1)/(t_2-t_1)\):\[\bar a_1=\frac{0.35\,v_D-0.31\,v_D}{0.12\,T}=\frac{+0.04\,v_D}{0.12\,T}=+0.33\;v_D\,T^{-1}\]\[\bar a_2=\frac{0.47\,v_D-0.62\,v_D}{0.12\,T}=\frac{-0.15\,v_D}{0.12\,T}=-1.2\;v_D\,T^{-1}\]\[\bar a_3=\frac{0.98\,v_D-0.98\,v_D}{0.13\,T}=\frac{0}{0.13\,T}=0\;v_D\,T^{-1}\]Comparing magnitudes, \(\displaystyle 1.2 > 0.33 > 0\): interval $\displaystyle 2$ wins, by roughly a factor of four over interval 1. This is the eye-check too — over three strips of the same width, the curve in strip $\displaystyle 2$ crosses far more vertical distance than in strip $\displaystyle 1$, and none at all in strip 3.Which interval has the greatest average speed? Average speed is the distance covered divided by the time, i.e. the mean height of the curve across the strip (the area under the curve divided by \(\displaystyle \Delta t\)). Strip $\displaystyle 1$ sits low on the graph (mean height \(\displaystyle \approx0.32\,v_D\)), strip $\displaystyle 2$ is halfway up (\(\displaystyle \approx0.54\,v_D\)), and strip $\displaystyle 3$ sits right on top of the tallest crest (\(\displaystyle \approx1.0\,v_D\)). So interval $\displaystyle 3$ has the greatest average speed — no arithmetic needed, it is simply the strip whose piece of curve is highest above the \(\displaystyle t\)-axis.Signs of \(\displaystyle v\) and \(\displaystyle a\). The graph is a speed graph, so it never goes below the axis; the particle never turns round. With the positive direction chosen along the motion, \(\displaystyle v>0\) in all three intervals. Accelerations at \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) and \(\displaystyle D\). Here we need the tangent, not a chord: \(\displaystyle a=dv/dt\). Each of these four points is a place where the curve is momentarily flat — \(\displaystyle A\) is on the level start of the curve, \(\displaystyle B\) and \(\displaystyle D\) are crests (maxima), \(\displaystyle C\) is a trough (minimum). At a maximum or a minimum the tangent to the curve is horizontal, so \(\displaystyle dv/dt = 0\) at each of them:\[a_A=a_B=a_C=a_D=0\]Answer: The average acceleration is greatest in magnitude in interval $\displaystyle 2$ (\(\displaystyle \bar a\approx-1.2\,v_D\,T^{-1}\), about four times interval $\displaystyle 1$'s \(\displaystyle +0.33\,v_D\,T^{-1}\)); the average speed is greatest in interval $\displaystyle 3$ (mean height \(\displaystyle \approx1.0\,v_D\), on the crest \(\displaystyle D\)). With the direction of motion taken positive, \(\displaystyle v>0\) in all three intervals, while \(\displaystyle a>0\) in interval $\displaystyle 1$, \(\displaystyle a<0\) in interval $\displaystyle 2$ (directed opposite to the motion), and \(\displaystyle a=0\) in interval 3. At the four marked points the tangent to the speed–time curve is horizontal, so \(\displaystyle a_A=a_B=a_C=a_D=0\).