Describing Motion Around Us Class 9 Science

20 Questions | 45 Marks

An athlete starts from rest and accelerates uniformly to a speed of 10 m/s in 5 seconds. She then runs at a constant speed for 10 seconds and finally applies brakes to stop in 6 seconds. Calculate the total distance travelled by the athlete. 5marks

A car moving with a velocity of 20 m/s applies brakes which cause a uniform deceleration of 4 m/s². Calculate the distance the car travels before coming to rest. 3marks

A train moving at a constant speed of 60 km/h enters a tunnel. The train takes 2 minutes to pass completely through the tunnel. What is the length of the tunnel? 3marks

A particle moves in a straight line. Its position-time graph is shown in Fig. 4.16a. What can you say about the velocities of objects A and B? 2marks

A bus travelling at 54 km/h applies brakes and comes to rest in 10 seconds. Calculate the acceleration of the bus and the distance travelled during this time. 3marks

Explain why it is important to maintain a safe distance from the vehicle ahead while driving. 2marks

A ball is thrown vertically upwards with a velocity of 20 m/s. It experiences a constant downward acceleration due to gravity (g = 9.8 m/s²). Calculate: (a) the maximum height reached by the ball, (b) the total time taken by the ball to return to its starting point. 5marks

What is the difference between distance and displacement? Under what condition is the magnitude of displacement equal to the distance travelled? 2marks

An object moves along a circular path of radius R. If it completes one revolution in time T, derive an expression for its average speed. 2marks

A car is moving on a straight highway at a velocity of 54 km/h. The driver presses the accelerator for 10 s, and the velocity increases to 72 km/h. Calculate the average acceleration during this interval. 3marks

Define uniform circular motion. Give two examples. 2marks

The velocity-time graph of a car is shown in Fig. 4.17(b). What can you conclude about the motion of the car? 1mark

A student runs from the ground floor to the fourth floor of a school building and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point. 3marks

Explain the term 'average acceleration'. Write its SI unit. 2marks

An object starts from rest and moves with uniform acceleration. If it travels 50 m in the first 2 seconds and 100 m in the first 4 seconds, what is the acceleration? 3marks

A circular track has a radius of 100 m. A cyclist travels at a constant speed and completes one revolution in 50 seconds. What is the average velocity of the cyclist during this time? 2marks

State the kinematic equations for motion with constant acceleration. 2marks

A particle is moving in a straight line with constant acceleration. Its velocity at a point A is 5 m/s and at point B is 10 m/s. If the distance between A and B is 20 m, calculate the acceleration. 3marks

Fig. 4.15 shows a position-time graph of a vehicle. What does this graph indicate about the motion of the vehicle? 2marks

What is instantaneous velocity? How does it differ from average velocity? 2marks

Answers

To calculate the total distance travelled, we need to find the distance travelled in each phase:

Phase 1 (Acceleration): Initial velocity $u = 0$ m/s, final velocity $v = 10$ m/s, time $t = 5$ s.
Using $v = u + a t$ , we get $10 = 0 + a \times 5 \Rightarrow a = 2$ m/s².
Using $s = u t + \frac{1}{2} a t^{2}$ , distance $s_{1} = 0 \times 5 + \frac{1}{2} \times 2 \times (5)^{2} = 25$ m.
Phase 2 (Constant Speed): Velocity $v = 10$ m/s, time $t = 10$ s.
Distance $s_{2} = velocity \times time = 10 m/s \times 10 s = 100$ m.
Phase 3 (Braking): Initial velocity $u = 10$ m/s, final velocity $v = 0$ m/s, time $t = 6$ s.
Using $v = u + a t$ , we get $0 = 10 + a \times 6 \Rightarrow a = - \frac{10}{6} = - \frac{5}{3}$ m/s².
Using $s = u t + \frac{1}{2} a t^{2}$ , distance $s_{3} = 10 \times 6 + \frac{1}{2} \times (- \frac{5}{3}) \times (6)^{2} = 60 + \frac{1}{2} \times (- \frac{5}{3}) \times 36 = 60 - 30 = 30$ m.
Total distance travelled $= s_{1} + s_{2} + s_{3} = 25 m + 100 m + 30 m = 155$ m. [5 marks]

Initial velocity

u = 20

m/s, acceleration

a = - 4

m/s², final velocity

v = 0

m/s.Using the kinematic equation

v^{2} = u^{2} + 2 a s

(0 m/s)^{2} = (20 m/s)^{2} + 2 \times (- 4 {m/s}^{2}) \times s

0 = 400 - 8 s

8 s = 400

$s = \frac{400}{8} = 50$ m. [3 marks]

Initial velocity

u = 60

km/h

= 60 \times \frac{1000}{3600}

m/s

= \frac{50}{3}

m/s.Time taken to pass through the tunnel

t = 2

minutes

= 2 \times 60 = 120

s.The distance covered by the train in 120 s is equal to the length of the tunnel plus the length of the train. However, the question asks for the length of the tunnel, implying we can consider the train's length negligible for this calculation or that the 'distance covered' refers to the length of the tunnel itself if the train's length is not given. Assuming we need to find the length of the tunnel based on the distance the train travels:Distance travelled = velocity

\times

timeDistance

= \frac{50}{3} m/s \times 120 s = 50 \times 40 = 2000

m.If the question implies that the train enters the tunnel and then exits it, and the time given is for the entire train to pass through it, then the distance travelled is indeed the length of the tunnel plus the length of the train. Without the train's length, we assume the question is asking for the distance covered by the train in that time.Length of tunnel = 2000 m or 2 km. 3marks

The position-time graph of object A is a straight line with a positive slope, indicating that object A is moving with a constant positive velocity. The position-time graph of object B is also a straight line with a positive slope, but it is steeper than that of A. This means object B is moving with a constant velocity that is higher than that of A. 2marks

Initial velocity

u = 54

km/h

= 54 \times \frac{1000}{3600}

m/s

= 15

m/s.Final velocity

v = 0

m/s (comes to rest).Time taken

t = 10

s.(a) Acceleration: Using

v = u + a t

0 = 15 + a \times 10

10 a = - 15

$a = - \frac{15}{10} = - 1.5$ m/s². The acceleration is $- 1.5$ m/s².
(b) Distance travelled: Using $v^{2} = u^{2} + 2 a s$ :

(0)^{2} = (15)^{2} + 2 \times (- 1.5) \times s

0 = 225 - 3 s

3 s = 225

$s = \frac{225}{3} = 75$ m. The distance travelled is 75 m. [3 marks]

Maintaining a safe distance is crucial for road safety. If the vehicle ahead suddenly brakes or slows down, a safe distance allows you enough time and space to react and apply your brakes without colliding with it. It also accounts for your own reaction time and the braking distance of your vehicle. 2marks

Given: Initial velocity

u = 20

m/s, acceleration due to gravity

a = - 9.8

m/s² (taking upward direction as positive).(a) To find the maximum height reached, we consider the point where the velocity becomes zero (

v = 0

m/s).Using

v^{2} = u^{2} + 2 a s

(0)^{2} = (20)^{2} + 2 \times (- 9.8) \times s_{m a x}

0 = 400 - 19.6 s_{m a x}

19.6 s_{m a x} = 400

$s_{m a x} = \frac{400}{19.6} \approx 20.41$ m. The maximum height reached is approximately 20.41 m.
(b) To find the total time taken to return to starting point, we consider the displacement to be zero ( $s = 0$ ).
Using $s = u t + \frac{1}{2} a t^{2}$ :

0 = 20 t + \frac{1}{2} (- 9.8) t^{2}

0 = 20 t - 4.9 t^{2}

t (20 - 4.9 t) = 0

This gives two solutions: $t = 0$ (initial point) and $20 - 4.9 t = 0$ .

4.9 t = 20

$t = \frac{20}{4.9} \approx 4.08$ s. The total time taken is approximately 4.08 seconds. [5 marks]

Distance is the total length of the path covered by an object. It is a scalar quantity.Displacement is the shortest straight-line distance between the initial and final positions of an object. It is a vector quantity, having both magnitude and direction.The magnitude of displacement is equal to the distance travelled when the object moves in a straight line without changing its direction. 2marks

The speed of the object is constant. Let the constant speed be

v

.The time taken to complete one revolution is

T

.The distance travelled in one revolution is the circumference of the circular path, which is

2 π R

.Using the formula: average speed

= \frac{total distance travelled}{time interval}

v = \frac{2 π R}{T}

. 2marks

Initial velocity

u = 54

km/h

= 54 \times \frac{1000}{3600}

m/s

= 15

m/s.Final velocity

v = 72

km/h

= 72 \times \frac{1000}{3600}

m/s

= 20

m/s.Time interval

t = 10

s.Average acceleration

a = \frac{v - u}{t}

a = \frac{20 m/s - 15 m/s}{10 s} = \frac{5 m/s}{10 s} = 0.5

m/s². 3marks

Uniform circular motion is the motion of an object in a circular path with constant speed.Examples:(i) The motion of the Earth around the Sun (approximately).(ii) The motion of an electron in a hydrogen atom (simplified model). 2marks

The velocity-time graph in Fig. 4.17(b) shows a straight line with a positive slope, indicating that the velocity of the car is increasing linearly with time. This means the car is moving with constant positive acceleration. 1mark

Let the height of each floor be

h = 3

m.The girl starts from the ground floor (let's say position 0).She goes to the fourth floor, so her displacement upwards is

4 h

.She then comes down to the second floor, so her displacement downwards is

2 h

.(i) Total vertical distance travelled:Distance to the fourth floor =

4 h

.Distance from the fourth floor to the second floor =

4 h - 2 h = 2 h

.Total distance travelled

= 4 h + 2 h = 6 h = 6 \times 3 m = 18

m.(ii) Displacement from the starting point:Starting position = Ground floor (0).Final position = Second floor.Displacement

= Final position - Initial position = 2 h - 0 = 2 h = 2 \times 3 m = 6

m upwards. 3marks

Average acceleration is defined as the change in velocity divided by the time interval during which the change occurs.Mathematically,

a_{a v g} = \frac{Δ v}{Δ t}

.The SI unit of average acceleration is metre per second squared (m/s²). 2marks

Let the acceleration be

a

.For the first 2 seconds, distance

s_{1} = 2

m and time

t_{1} = 2

s.Using

s = u t + \frac{1}{2} a t^{2}

, and

u = 0

(starts from rest):

2 = 0 \times 2 + \frac{1}{2} \times a \times (2)^{2}

2 = \frac{1}{2} \times a \times 4

$2 = 2 a \Rightarrow a = 1$ m/s².
Now let's check this with the data for the first 4 seconds.
Distance $s_{2} = 100$ m and time $t_{2} = 4$ s.
Using $s = u t + \frac{1}{2} a t^{2}$ , and $u = 0$ :

100 = 0 \times 4 + \frac{1}{2} \times a \times (4)^{2}

100 = \frac{1}{2} \times a \times 16

$100 = 8 a \Rightarrow a = \frac{100}{8} = 12.5$ m/s².
The values of acceleration are different, which means there might be an issue with the question data or interpretation. Let's re-examine the problem statement. It seems the problem implies two different scenarios or there's a contradiction. If we assume the question implies that for a certain motion, both conditions must hold, then it's contradictory.

Let's assume the question means:
Scenario 1: An object starts from rest and accelerates uniformly. It travels 50 m in the first 2 seconds. Find acceleration.

s = u t + \frac{1}{2} a t^{2}

50 = 0 (2) + \frac{1}{2} a (2)^{2}

$50 = 2 a \Rightarrow a = 25$ m/s².

Scenario 2: An object starts from rest and accelerates uniformly. It travels 100 m in the first 4 seconds. Find acceleration.

s = u t + \frac{1}{2} a t^{2}

100 = 0 (4) + \frac{1}{2} a (4)^{2}

$100 = 8 a \Rightarrow a = 12.5$ m/s².

If the question is asking for a single acceleration that satisfies both, it's not possible with the given numbers.

Let's assume there is a typo and interpret the question as:
"An object starts from rest and accelerates uniformly. In the first 2 seconds, it travels a distance $s_{1}$ . In the first 4 seconds, it travels a total distance of 100 m. If the distance travelled in the first 2 seconds is 50m, find the acceleration." This still leads to a contradiction.

Let's assume the question meant:
"An object starts from rest and accelerates uniformly. The distance covered in the first 2 seconds is 50m. The distance covered in the next 2 seconds (from t=2s to t=4s) is 100m. Find the acceleration."
Let acceleration be $a$ .
Distance in first 2 sec: $s_{1} = \frac{1}{2} a (2)^{2} = 2 a$ .
Distance in first 4 sec: $s_{1 + 2} = \frac{1}{2} a (4)^{2} = 8 a$ .
Distance in the next 2 sec = $s_{1 + 2} - s_{1} = 8 a - 2 a = 6 a$ .
Given $s_{1} = 50$ m, so $2 a = 50 \Rightarrow a = 25$ m/s².
Given distance in the next 2 seconds is 100 m. So, $6 a = 100 \Rightarrow a = \frac{100}{6} = \frac{50}{3} \approx 16.67$ m/s².
This interpretation also leads to a contradiction.

Assuming the intended question was:
"An object starts from rest and accelerates uniformly. It travels 50 m in the first 2 seconds. Find the acceleration. Then, if it travels 100 m in the next 2 seconds (from t=2s to t=4s), find the acceleration." This leads to two different accelerations.

Let's consider the most likely interpretation for a typical exam question where 'in the first X seconds' means the total distance from t=0:
If the question states "it travels 50 m in the first 2 seconds AND 100 m in the first 4 seconds", then the only way to get a single acceleration is if these are correct.
From first 2 sec: $s_{1} = 50$ m, $t_{1} = 2$ s, $u = 0$ .
$50 = 0 + \frac{1}{2} a (2)^{2} \Rightarrow 50 = 2 a \Rightarrow a = 25$ m/s².
From first 4 sec: $s_{2} = 100$ m, $t_{2} = 4$ s, $u = 0$ .
$100 = 0 + \frac{1}{2} a (4)^{2} \Rightarrow 100 = 8 a \Rightarrow a = 12.5$ m/s².
There is a contradiction in the given data. Assuming the data for the first 4 seconds is correct and is intended to be used for calculation:
Acceleration $a = 12.5$ m/s². [3 marks, acknowledging data inconsistency]

The cyclist starts from rest and completes one revolution in 50 seconds.Initial position = Final position (after one revolution).Displacement = 0 m.Average velocity =

\frac{Displacement}{Time interval} = \frac{0 m}{50 s} = 0

m/s. 2marks

The kinematic equations for motion with constant acceleration are:(i)

v = u + a t

(ii)

s = u t + \frac{1}{2} a t^{2}

(iii)

v^{2} = u^{2} + 2 a s

where

u

is initial velocity,

v

is final velocity,

a

is acceleration,

t

is time interval, and

s

is displacement. 2marks

Initial velocity

u = 5

m/s, final velocity

v = 10

m/s, displacement

s = 20

m.Using the kinematic equation

v^{2} = u^{2} + 2 a s

(10 m/s)^{2} = (5 m/s)^{2} + 2 \times a \times 20 m

100 = 25 + 40 a

100 - 25 = 40 a

75 = 40 a

$a = \frac{75}{40} = \frac{15}{8} = 1.875$ m/s². [3 marks]

The position-time graph in Fig. 4.15 is a horizontal line parallel to the time axis. This indicates that the position of the vehicle is constant with respect to time. Therefore, the vehicle is at rest at a position of 40 m from the origin. 2marks

Instantaneous velocity is the velocity of an object at a particular instant of time. It is the magnitude of the instantaneous rate of change of position.Average velocity is the displacement of an object divided by the time interval during which the displacement occurs. It represents the overall change in position over a period of time.The key difference is that instantaneous velocity refers to a single moment, while average velocity considers a time interval. If the velocity is changing, instantaneous velocity can be very different from average velocity. For example, if an object moves forward and then backward, its average velocity might be zero, but its instantaneous velocity at various points can be non-zero. 2marks

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Describing Motion Around Us — Important Questions

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