NCERT Solutions · Class 9 Science Describing Motion Around Us

His journey has four legs, each of \(\displaystyle 250\ \text{m} \): home → shop, shop → home (for the bag), home → shop again, shop → home.

\(\displaystyle \text{total distance} = 4 \times 250\ \text{m} = 1000\ \text{m} \)

He finishes exactly where he began, so the net change in position is zero: \(\displaystyle s = 0\ \text{m} \).

This is the standard case of a large distance with zero displacement.

(i) Total vertical distance travelled \(\displaystyle = 18\ \text{m} \).

Going up, ground floor to fourth floor is $\displaystyle 4$ floors: \(\displaystyle 4 \times 3\ \text{m} = 12\ \text{m} \).

Coming down, fourth floor to second floor is $\displaystyle 2$ floors: \(\displaystyle 2 \times 3\ \text{m} = 6\ \text{m} \).

\(\displaystyle \text{total distance} = 12\ \text{m} + 6\ \text{m} = 18\ \text{m} \)

(ii) Displacement \(\displaystyle = 6\ \text{m} \), in the upward direction.

The student starts on the ground floor and ends on the second floor, a net rise of \(\displaystyle 2 \times 3\ \text{m} = 6\ \text{m} \); the trip up to the fourth floor and back does not add to it.

Yes — her scooter can be accelerating, and it is accelerating whenever she is turning. A speedometer reports only the magnitude of her velocity; the direction of that velocity can be changing all the while, and a change in direction alone is already a change in velocity.

What a steady needle actually fixes — the Threads of Curiosity box beside Section $\displaystyle 4.1.4$ (the Average acceleration page) says the speedometer reading is nearly, though not exactly, the magnitude of the velocity at that instant, while it is the way the tyres are pointing that gives the direction of the velocity. So the needle pins down one half of the velocity and says nothing at all about the other half.

Why half is not enough — Eq. (4.3a) defines \(\displaystyle \text{average acceleration} = \frac{\text{change in velocity}}{\text{time interval}} \), and Section $\displaystyle 4.1.4$ states in as many words that this change can come from the magnitude of the velocity, from its direction, or from both. That same paragraph forward-refers to Section $\displaystyle 4.4$ for an example in which the speed stays constant and only the direction changes — which is precisely the situation this question describes.

How, concretely: she must be riding along a curve — a bend in the road, a roundabout, a circular turn, i.e. uniform circular motion. Section $\displaystyle 4.4.1$ reaches it by counting turns: an athlete on a rectangular track changes direction $\displaystyle 4$ times per lap (Fig. 4.23a), on a hexagonal track $\displaystyle 6$ times (Fig. 4.23b), and as the number of sides is increased without limit the track becomes a circle and the direction of the velocity changes continuously (Fig. 4.23c). At each point the velocity lies along the tangent to the path (Fig. $\displaystyle 4.25$), so it points a new way at every instant while its magnitude never moves.

Section $\displaystyle 4.4.1$ then states the conclusion the question is built on: uniform circular motion is accelerated motion, for the single reason that the direction of the velocity keeps changing; and the Note beside it adds that in everyday life we call a vehicle "accelerating" only when its speed changes and routinely fail to notice acceleration that is a change of direction alone. Her scooter is exactly the case that Note is about.

When she would not be accelerating — only if the road is straight as well as her speed steady. Then the initial and final velocities agree in magnitude and direction, and Eq. (4.3c) gives \(\displaystyle a = \frac{v - u}{t} = \frac{10\ \mathrm{m\ s^{-1}} - 10\ \mathrm{m\ s^{-1}}}{t} = 0\ \mathrm{m\ s^{-2}} \) for any time interval (the \(\displaystyle 10\ \mathrm{m\ s^{-1}} \) is only there to show the substitution; any steady reading gives the same zero). The Note in Section $\displaystyle 4.1.4$ makes the same point with a bus on a straight highway — moving fast, yet with zero acceleration.

Marked as reasoning rather than the book's content: the chapter establishes only that this acceleration is non-zero; it never says which way it points, and nothing in Chapter $\displaystyle 4$ mentions an acceleration directed towards the centre of the circle — so a Grade $\displaystyle 9$ answer should not supply a direction here. Similarly, the speedometer box's "nearly, but not exactly" means a perfectly steady needle is a very good indicator of constant speed rather than a strict guarantee of it.

The printed key does not fit this question. The book's Answer Key gives, under Chapter $\displaystyle 4$ → Revise, Reflect, Refine, item 3. Yes, Yes, different — three answers. Question $\displaystyle 3$ as printed is one sentence: a single yes/no ("Is it possible for her scooter to be accelerating") followed by "if so, how?". A how cannot be answered "Yes", and there is nothing in the question that can be answered "different". A three-part entry of exactly that shape does fit Pause and Ponder question $\displaystyle 3$ of the same chapter — the ball on the inclined track of Fig. $\displaystyle 4.6$ — which asks three things in a row: is the motion straight-line motion, can it be depicted on a horizontal line as in Fig. $\displaystyle 4.3$, and are the total distance travelled and the magnitude of displacement equal or different. The Answer Key's Pause and Ponder list for Chapter $\displaystyle 4$ prints only item $\displaystyle 4$, with no item $\displaystyle 3$; and the key does print yes/no answers to Pause and Ponder items in other chapters (Chapter $\displaystyle 7$'s list opens "1. No"), so that gap is not a policy of skipping non-numerical answers. The entry looks set against the wrong list.

So, plainly: the book's printed Answer Key gives "Yes, Yes, different" for this question, which does not agree with the question as printed on the Revise, Reflect, Refine page or with Section $\displaystyle 4.4.1$ read as an answer to it. The chapter's own text settles it the other way — Section $\displaystyle 4.1.4$ counts a change of direction as a change of velocity, and Section $\displaystyle 4.4.1$ declares uniform circular motion accelerated for that reason alone — so the answer is Yes: a scooter held at a constant speedometer reading is accelerating whenever it is going round a bend, and is not accelerating only if it is also going straight. Of the key's three words, only the first one survives.

Given: \(\displaystyle u = 0\ \text{m s}^{-1} \) (starts from rest), \(\displaystyle v = 24\ \text{m s}^{-1} \), \(\displaystyle t = 6\ \text{s} \).

Using Eq. (4.3c): \(\displaystyle a = \frac{v - u}{t} = \frac{24\ \text{m s}^{-1} - 0\ \text{m s}^{-1}}{6\ \text{s}} = 4\ \text{m s}^{-2} \)

The magnitude of velocity is increasing, so the acceleration acts along the direction of motion.

Using Eq. (4.4b): \(\displaystyle s = ut + \frac{1}{2}at^{2} = 0 + \frac{1}{2} \times 4\ \text{m s}^{-2} \times (6\ \text{s})^{2} = 72\ \text{m} \)

Given: \(\displaystyle u = 28\ \text{m s}^{-1} \), \(\displaystyle v = 0\ \text{m s}^{-1} \) (it stops), \(\displaystyle s = 98\ \text{m} \).

Using Eq. (4.4c): \(\displaystyle v^{2} = u^{2} + 2as \Rightarrow 0 = (28\ \text{m s}^{-1})^{2} + 2 \times a \times 98\ \text{m} \)

\(\displaystyle a = \frac{-784\ \text{m}^{2}\text{s}^{-2}}{196\ \text{m}} = -4\ \text{m s}^{-2} \)

The minus sign says the acceleration is directed opposite to the velocity — the motorbike is slowing down.

Using Eq. (4.4a): \(\displaystyle 0 = 28\ \text{m s}^{-1} + (-4\ \text{m s}^{-2}) \times t \Rightarrow t = \frac{28}{4}\ \text{s} = 7\ \text{s} \)

Both graphs in Fig. $\displaystyle 4.27$ are straight lines, so each object moves with its own constant velocity throughout the motion.

On a position-time graph the velocity is the slope of the line. The two lines have visibly different steepness, so the two constant velocities are different numbers — and being constant, they stay different at every instant.

Where the two lines meet, A and B have the same position, i.e. they are side by side on their parallel tracks and one overtakes the other. Equal position is not equal velocity.

(i) is true: both objects have the same initial position and the same final position, so both have the same displacement over the same \(\displaystyle 10\ \text{s} \); reading the graph, \(\displaystyle v_{av} = \frac{10\ \text{m}}{10\ \text{s}} = 1\ \text{m s}^{-1} \) for each.

(ii) is true: neither graph ever falls, so neither object turns back. For one-way motion the distance travelled equals the magnitude of displacement, so each covers \(\displaystyle 10\ \text{m} \) and \(\displaystyle \text{average speed} = \frac{10\ \text{m}}{10\ \text{s}} = 1\ \text{m s}^{-1} \) for both.

(iii) is false: A does not cover a shorter distance than B — both cover the same \(\displaystyle 10\ \text{m} \).

(iv) is false: B is indeed slower than A over some stretches, but it is faster over others, and over the full \(\displaystyle 10\ \text{s} \) the two average out to the same speed.

What really differs between A and B is their velocity at individual instants, not their averages over the whole interval.

Convert the speeds: \(\displaystyle u = 54\ \text{km h}^{-1} = 54 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 15\ \text{m s}^{-1} \) and \(\displaystyle v = 36\ \text{km h}^{-1} = 10\ \text{m s}^{-1} \), with \(\displaystyle t = 36\ \text{s} \).

Using Eq. (4.3c): \(\displaystyle a = \frac{10\ \text{m s}^{-1} - 15\ \text{m s}^{-1}}{36\ \text{s}} = -\frac{5}{36}\ \text{m s}^{-2} \approx -0.14\ \text{m s}^{-2} \) — negative, because he is slowing down.

Using Eq. (4.4b): \(\displaystyle s = ut + \frac{1}{2}at^{2} = 15 \times 36 + \frac{1}{2} \times \left(-\frac{5}{36}\right) \times (36)^{2} \)

\(\displaystyle s = 540\ \text{m} - 90\ \text{m} = 450\ \text{m} \)

The \(\displaystyle 40\ \text{km h}^{-1} \) figure on the sign is not used in the calculation — it is only the reason he slows down.

Total distance travelled \(\displaystyle = 310\ \text{m} \). Treat the journey as three stages and add the distances.

Stage $\displaystyle 1$, speeding up: \(\displaystyle u = 0 \), \(\displaystyle v = 20\ \text{m s}^{-1} \), \(\displaystyle t = 5\ \text{s} \), so \(\displaystyle a = \frac{20 - 0}{5} = 4\ \text{m s}^{-2} \) and \(\displaystyle s_{1} = \frac{1}{2} \times 4\ \text{m s}^{-2} \times (5\ \text{s})^{2} = 50\ \text{m} \).

Stage $\displaystyle 2$, constant velocity: \(\displaystyle s_{2} = 20\ \text{m s}^{-1} \times 10\ \text{s} = 200\ \text{m} \).

Stage $\displaystyle 3$, braking: \(\displaystyle u = 20\ \text{m s}^{-1} \), \(\displaystyle v = 0 \), \(\displaystyle t = 6\ \text{s} \), so \(\displaystyle a = \frac{0 - 20}{6} = -\frac{10}{3}\ \text{m s}^{-2} \) and \(\displaystyle s_{3} = 20 \times 6 + \frac{1}{2} \times \left(-\frac{10}{3}\right) \times (6)^{2} = 120\ \text{m} - 60\ \text{m} = 60\ \text{m} \).

\(\displaystyle \text{total} = 50\ \text{m} + 200\ \text{m} + 60\ \text{m} = 310\ \text{m} \)

Convert: \(\displaystyle u = 36\ \text{km h}^{-1} = 36 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 10\ \text{m s}^{-1} \).

Reaction stretch — the brakes are not yet on, so the bus keeps its constant velocity for \(\displaystyle 0.5\ \text{s} \): \(\displaystyle s_{1} = 10\ \text{m s}^{-1} \times 0.5\ \text{s} = 5\ \text{m} \).

Braking stretch — using Eq. (4.4c) with \(\displaystyle v = 0 \) and \(\displaystyle a = -2.5\ \text{m s}^{-2} \): \(\displaystyle 0 = (10)^{2} + 2 \times (-2.5) \times s_{2} \Rightarrow s_{2} = \frac{100}{5} = 20\ \text{m} \).

Total stopping distance \(\displaystyle = 5\ \text{m} + 20\ \text{m} = 25\ \text{m} \), and \(\displaystyle 25\ \text{m} < 30\ \text{m} \).

Note how much the reaction time costs: a fifth of the whole stopping distance is covered before the brake is even touched.