Describing Motion Around Us — NCERT Solutions

The child travels along the curved path ABC. So the distance travelled is the length of the path from A to B to C.

The displacement is the straight-line distance from the original position to the final position, i.e. AC.

In one revolution, the child goes once around the circle, so the distance travelled is the circumference of the circle. If the radius is $R$, this distance is $2\pi R$.

For one revolution, the child travels a distance of $2\pi R$ in time $T$. So,

$$v_{av}=\frac{\text{distance travelled}}{\text{time taken}}=\frac{2\pi R}{T}$$

Since the child returns to the original position after one revolution, the displacement is 0. Therefore the average velocity over the time interval $T$ is also

$$\text{average velocity}=\frac{0}{T}=0$$

Yes. In uniform circular motion, the speed is constant but the direction of velocity changes continuously at every point on the circle.

If the ring is lifted while the marble is moving, the marble will no longer keep moving in a circle. It will move in a straight line in the direction it had at the instant the ring was removed.

After the ring is picked up, the marble does not continue in circular motion. It moves in a straight line along the direction of motion it had at that instant.

The shop is 250 m from home.

• Home to shop = 250 m
• Shop to home = 250 m
• Home to shop again = 250 m
• Shop to home again = 250 m

So, total distance travelled:

$$250+250+250+250=1000\text{ m}$$

Since he finally comes back to home, his final position is the same as his starting position.
So, displacement = 0 m.

Ground floor to fourth floor means moving up 4 floors.
Each floor is 3 m high.

(i) Total vertical distance travelled

From ground to fourth floor:

$$4\times 3=12\text{ m}$$

Then coming down to second floor means moving down 2 floors:

$$2\times 3=6\text{ m}$$

Total vertical distance:

$$12+6=18\text{ m}$$

(ii) Displacement

Starting at ground floor and ending at second floor means net change is 2 floors upward:

$$2\times 3=6\text{ m}$$

So displacement is 6 m upward. If downward is taken as positive in some convention, it would be negative; here the natural direction is upward.

Yes. Even if the speedometer reading is constant, the scooter can still be accelerating if its direction of motion changes. Acceleration depends on change in velocity, and velocity includes direction.

Initial velocity $u=0$ since the car starts from rest.
Final velocity $v=24\,\mathrm{m\,s^{-1}}$
Time $t=6\,\mathrm{s}$

Average acceleration

$$a=\frac{v-u}{t}=\frac{24-0}{6}=4\,\mathrm{m\,s^{-2}}$$

Distance travelled

Using

$$s=ut+\frac12 at^2$$

$$s=0\times 6+\frac12\times 4\times 6^2$$

$$s=2\times 36=72\,\mathrm{m}$$

Given:

• initial velocity $u=28\,\mathrm{m\,s^{-1}}$
• final velocity $v=0$ since it stops
• displacement $s=98\,\mathrm{m}$

Use

$$v^2=u^2+2as$$

Substitute values:

$$0^2=28^2+2a(98)$$
$$0=784+196a$$
$$196a=-784$$
$$a=-4\,\mathrm{m\,s^{-2}}$$

Now use

$$v=u+at$$

$$0=28+(-4)t$$
$$4t=28$$
$$t=7\,\mathrm{s}$$

From a position-time graph, velocity is given by the slope of the graph. So, to know whether objects A and B ever have equal velocity, we must compare the slopes of their graphs at the same instant.

If the graphs have the same slope at any instant, then their velocities are equal at that instant. If the slopes are never equal, then their velocities are never equal. The answer depends on the shape of the given graph in Fig. 4.27.

The correct option is (i).

Since both objects have the same initial and final positions over the 10 s interval, their average velocity is equal. The other statements are not necessarily true from the graph alone.

Given:

• initial speed $u=54\,\mathrm{km\,h^{-1}}=15\,\mathrm{m\,s^{-1}}$
• final speed $v=36\,\mathrm{km\,h^{-1}}=10\,\mathrm{m\,s^{-1}}$
• time $t=36\,\mathrm{s}$

Since acceleration is constant, displacement is

$$s=\frac{1}{2}(u+v)t$$

$$s=\frac{1}{2}(15+10)\times 36$$

$$s=\frac{25}{2}\times 36=25\times 18=450\,\mathrm{m}$$

So the distance travelled is 450 m. If the answer expected from the chapter’s direct use of the kinematic relation is needed, the computed value is 450 m.

Break the motion into three parts.

1) Accelerating from rest to 20 m s⁻¹ in 5 s

$$u=0,\ v=20,\ t=5$$

Distance:

$$s_1=\frac{1}{2}(u+v)t=\frac{1}{2}(0+20)\times 5=50\,\mathrm{m}$$

2) Constant speed 20 m s⁻¹ for 10 s

$$s_2=vt=20\times 10=200\,\mathrm{m}$$

3) Braking to stop in 6 s

$$u=20,\ v=0,\ t=6$$

Distance:

$$s_3=\frac{1}{2}(u+v)t=\frac{1}{2}(20+0)\times 6=60\,\mathrm{m}$$

Total distance

$$s=s_1+s_2+s_3=50+200+60=310\,\mathrm{m}$$

So the total distance travelled is 310 m.

First convert the speed:

$$36\,\mathrm{km\,h^{-1}}=10\,\mathrm{m\,s^{-1}}$$

Distance covered during reaction time

Reaction time = 0.5 s

$$s_1=vt=10\times 0.5=5\,\mathrm{m}$$

Remaining distance to obstacle:

$$30-5=25\,\mathrm{m}$$

Braking distance

Using

$$v^2=u^2+2as$$

Here $u=10\,\mathrm{m\,s^{-1}}$, $v=0$, $a=-2.5\,\mathrm{m\,s^{-2}}$.

$$0=10^2+2(-2.5)s$$
$$0=100-5s$$
$$s=20\,\mathrm{m}$$

Total stopping distance:

$$5+20=25\,\mathrm{m}$$

Since the obstacle is 30 m away, the bus stops before reaching it. So the answer is Yes.

An object kept on the Earth can be considered at rest with respect to the Earth’s surface if its position does not change relative to the Earth.

But the Earth itself is moving around the Sun, so with respect to the Sun, the object is also moving. Therefore, rest and motion depend on the reference point. So the object is at rest relative to the Earth, but not at rest relative to the Sun.

The distance travelled from a velocity-time graph is equal to the area under the graph. So, to estimate the distance from Fig. 4.31, we would calculate the area enclosed by the graph and the time axis. The chapter explains this method for velocity-time graphs. The method is: distance = area under the velocity-time graph.

We divide the motion into two parts.

1) Constant velocity for 2 minutes

Velocity $v=6\,\mathrm{m\,s^{-1}}$
Time $t=2\text{ min}=120\text{ s}$

Distance:

$$s_1=vt=6\times 120=720\,\mathrm{m}$$

2) Accelerated motion for 6 s

Initial velocity $u=6\,\mathrm{m\,s^{-1}}$
Acceleration $a=1\,\mathrm{m\,s^{-2}}$
Time $t=6\,\mathrm{s}$

Distance:

$$s_2=ut+\frac12 at^2$$

$$s_2=6\times 6+\frac12\times 1\times 6^2$$

$$s_2=36+18=54\,\mathrm{m}$$

Total displacement

$$s=s_1+s_2=720+54=774\,\mathrm{m}$$

Car A

Starts from rest, reaches $5\,\mathrm{m\,s^{-1}}$ in 5 s.
So acceleration:

$$a_A=\frac{5-0}{5}=1\,\mathrm{m\,s^{-2}}$$

Velocity-time graph is a straight line from $(0,0)$ to $(5,5)$.

Displacement in 5 s = area under graph:

$$s_A=\frac12 \times 5 \times 5=12.5\,\mathrm{m}$$

Car B

Starts from rest, reaches $3\,\mathrm{m\,s^{-1}}$ in 10 s.
So acceleration:

$$a_B=\frac{3-0}{10}=0.3\,\mathrm{m\,s^{-2}}$$

Velocity-time graph is a straight line from $(0,0)$ to $(10,3)$.

Displacement in 10 s = area under graph:

$$s_B=\frac12 \times 10 \times 3=15\,\mathrm{m}$$

Comparison

• Car A: displacement in 5 s = 12.5 m
• Car B: displacement in 10 s = 15 m

To plot both on the same graph, mark time on the X-axis and velocity on the Y-axis, then join the points with straight lines from the origin.

For the tip of the minute hand:

• In 1 hour, it makes 1 complete revolution.
• So in 1.5 hours from 6:00 PM to 7:30 PM, it makes 1.5 revolutions.
• The length of the minute hand is 7 cm, so the radius of the circular path is 7 cm.

(i) Distance travelled

Distance in one revolution = circumference = $2\pi r = 2\pi \times 7 = 14\pi$ cm

In 1.5 revolutions:
$$\text{distance} = 1.5 \times 14\pi = 21\pi \text{ cm}$$
Using $\pi = \frac{22}{7}$:
$$21\pi = 21 \times \frac{22}{7} = 66 \text{ cm}$$

(ii) Displacement

From 6:00 to 7:30, the minute hand moves from 12 to 6, i.e. to the opposite point on the circle, so the displacement is the diameter:
$$2r = 2 \times 7 = 14 \text{ cm}$$

(iii) Speed

$$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{66}{1.5\times 3600} \text{ cm s}^{-1}$$
$$= \frac{66}{5400} = \frac{11}{900} \text{ cm s}^{-1}$$

(iv) Velocity

$$\text{velocity} = \frac{\text{displacement}}{\text{time}} = \frac{14}{5400} \text{ cm s}^{-1}$$

Its direction is along the straight line joining the initial and final positions of the tip, from the starting point towards the ending point.

Note: Since this is a circular motion, speed is not zero, but velocity is a vector and depends on displacement.

From the graph for 0 s to 120 s, the cyclist’s velocity is constant at 6 m s$^{-1}$ for the first 60 s and then decreases linearly from 6 m s$^{-1}$ to 0 in the next 60 s.

Displacement

Displacement = area under the velocity-time graph.

1. Rectangle part for first 60 s:

$$\text{area} = 6 \times 60 = 360 \text{ m}$$

2. Triangle part for next 60 s:

$$\text{area} = \frac{1}{2} \times 60 \times 6 = 180 \text{ m}$$

Total displacement:
$$360 + 180 = 540 \text{ m}$$

Average acceleration

$$a = \frac{v-u}{t}$$
Here,
$$u = 6 \text{ m s}^{-1},\quad v = 0,\quad t = 60 \text{ s}$$
So,
$$a = \frac{0-6}{60} = -0.1 \text{ m s}^{-2}$$

So the displacement is 540 m and the average acceleration is $-0.1\ \text{m s}^{-2}$.

The car moves in two parts.

1) Constant velocity for 2 minutes

$$v = 6\ \text{m s}^{-1},\quad t = 2\ \text{min} = 120\ \text{s}$$
Displacement:
$$s_1 = vt = 6 \times 120 = 720\ \text{m}$$

2) Accelerates for 6 s

Initial velocity:
$$u = 6\ \text{m s}^{-1}$$
Acceleration:
$$a = 1\ \text{m s}^{-2}$$
Time:
$$t = 6\ \text{s}$$
Using
$$s = ut + \frac{1}{2}at^2$$
$$s_2 = 6\times 6 + \frac{1}{2}\times 1 \times 6^2$$
$$s_2 = 36 + 18 = 54\ \text{m}$$

Total displacement

$$s = s_1 + s_2 = 720 + 54 = 774\ \text{m}$$

However, the chapter text for this style of problem usually assumes the second part is in the same straight line and asks for total displacement from the velocity-time graph. Using that, the displacement is the area under the graph: rectangle + trapezium.

Since velocity increases from 6 to 12 m s$^{-1}$ in 6 s:
$$s_2 = \frac{1}{2}(u+v)t = \frac{1}{2}(6+12)\times 6 = 54\ \text{m}$$
So total displacement remains:
$$720 + 54 = 774\ \text{m}$$

Final answer: 774 m

From the kinematic equations in the chapter:

$$v = u + at$$
$$s = ut + \frac{1}{2}at^2$$

The chapter asks to derive the remaining two equations:

1) Derive $s = vt - \frac{1}{2}at^2$

From
$$v = u + at$$
we get
$$u = v - at$$
Substitute this in
$$s = ut + \frac{1}{2}at^2$$
So,
$$s = (v-at)t + \frac{1}{2}at^2$$
$$s = vt - at^2 + \frac{1}{2}at^2$$
$$s = vt - \frac{1}{2}at^2$$

2) Derive $s = \frac{1}{2}(u+v)t$

From
$$v = u + at$$
we get
$$at = v-u$$
Substitute in
$$s = ut + \frac{1}{2}at^2$$
$$s = ut + \frac{1}{2}t(v-u)$$
$$s = ut + \frac{1}{2}vt - \frac{1}{2}ut$$
$$s = \frac{1}{2}ut + \frac{1}{2}vt$$
$$s = \frac{1}{2}(u+v)t$$

These are the two remaining kinematic equations.

The shape of a position-time graph tells us the nature of motion:

• A straight line position-time graph indicates constant velocity.
• A curved position-time graph indicates that the velocity is not constant, so the object is in accelerated motion.
• A straight line parallel to the time axis means the object is at rest.

So, by looking at the shape of the graph, we can tell whether the object is moving with constant velocity, changing velocity, or is stationary.

From a position-time graph, we can obtain:

• the position of the object at each instant of time,
• the average velocity by finding the slope of the graph.

So, the graph tells us both the object's position and how fast its position is changing.

When the driver presses the accelerator:

Given from the chapter:

• initial velocity, $u = 36\ \text{km h}^{-1} = 10\ \text{m s}^{-1}$
• final velocity, $v = 54\ \text{km h}^{-1} = 15\ \text{m s}^{-1}$
• time interval, $t = 10\ \text{s}$

Using
$$a = \frac{v-u}{t}$$
$$a = \frac{15-10}{10} = \frac{5}{10} = 0.5\ \text{m s}^{-2}$$

Since velocity is increasing, the acceleration is in the same direction as velocity.

When the driver presses the brake:

Given from the chapter:

• initial velocity, $u = 54\ \text{km h}^{-1} = 15\ \text{m s}^{-1}$
• final velocity, $v = 0\ \text{m s}^{-1}$
• time interval, $t = 5\ \text{s}$

Using
$$a = \frac{v-u}{t}$$
$$a = \frac{0-15}{5} = \frac{-15}{5} = -3\ \text{m s}^{-2}$$

The negative sign shows that the acceleration is opposite to the direction of motion, i.e. it is deceleration.