Class 9 Physics Motion Solved Numerical Problems

Formulas from Chapter 7 (Motion)

Formula	Description
v = u + at	Velocity-Time relation. Final velocity after time t.
s = ut + (1/2)at²	Position-Time relation. Distance covered in time t.
v² = u² + 2as	Relation between velocity, acceleration, and distance.
v = s/t	Speed formula. Average speed when distance s and time t are known.
vavg = (u + v)/2	Average velocity in case of uniform acceleration.
a = (v - u)/t	Acceleration formula. Change in velocity per unit time.
s = vt - (1/2)at²	Alternate position-time relation.
v = (2πr)/t	Velocity for uniform circular motion, where r is radius, t is time.

Explanation and Derivations of the Formulas

1. v = u + at (Velocity-Time Relation)

Explanation: This equation gives the final velocity v of an object starting with initial velocity u under uniform acceleration a over time t.

Derivation: From acceleration definition, a = (v - u) / t, rearranging gives v = u + at.

2. s = ut + (1/2)at² (Position-Time Relation)

Explanation: This equation gives the distance s covered by an object starting with initial velocity u and uniform acceleration a over time t.

Derivation: The average velocity over t is vavg = (u + v)/2. Using s = vavg × t and substituting v = u + at into vavg, we get:
s = ut + (1/2)at².

3. v² = u² + 2as (Position-Velocity Relation)

Explanation: This relates the square of the final velocity v to the initial velocity u, acceleration a, and distance s traveled.

Derivation: From s = ut + (1/2)at² and v = u + at, eliminate t: v² = u² + 2as.

4. v = s/t (Speed Formula)

Explanation: Speed is the distance covered per unit time. Valid for uniform motion.

5. vavg = (u + v)/2 (Average Velocity)

Explanation: Average velocity is the mean of initial and final velocities when acceleration is uniform.

6. a = (v - u)/t (Acceleration Formula)

Explanation: Acceleration is the rate of change of velocity over time.

7. s = vt - (1/2)at² (Alternate Position-Time Relation)

Explanation: This version calculates the distance s using the final velocity v and acceleration a.

Derivation: Substitute t = (v - u)/a into s = ut + (1/2)at².

8. v = (2πr)/t (Velocity for Uniform Circular Motion)

Explanation: For an object moving in a circle of radius r with period t, its speed is the circumference divided by time.

Class 9 Physics Motion Solved Numerical Problems from NCERT Book

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Solution:

Distance Covered:
Diameter of the circular track, d = 200 m
Radius of the circular track, r = d/2 = 100 m
Circumference of the circular track, C = 2 × π × r = 2 × π × 100 = 200π m
Time taken to complete one round, t₁ = 40 s

Total time, t₂ = 2 minutes 20 s = 140 s

Number of rounds completed in 140 s, n = t₂/t₁ = 140/40 = 3.5
Distance covered in 3.5 rounds, d = 3.5 × 200π = 700π m = 2199.1 m

Displacement:
Since the athlete completes 3.5 rounds, he ends up at the opposite point on the circular track.
The displacement is the diameter of the track, which is 200 m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Solution:

From A to B:
Distance \( d_{AB} = 300 \, \text{m} \)
Time \( t_{AB} = 2 \, \text{minutes} \, 30 \, \text{seconds} = 150 \, \text{s} \)
Average speed \( v_{\text{avg}, AB} = \frac{d_{AB}}{t_{AB}} = \frac{300}{150} = 2 \, \text{m/s} \)

Since he jogs in a straight line from A to B, the average velocity \( v_{AB} = v_{\text{avg}, AB} = 2 \, \text{m/s} \)

From A to C:
Distance \( d_{AC} = 300 + 100 = 400 \, \text{m} \)
Time \( t_{AC} = 150 + 60 = 210 \, \text{s} \)
Average speed \( v_{\text{avg}, AC} = \frac{d_{AC}}{t_{AC}} = \frac{400}{210} = 1.9 \, \text{m/s} \)
Displacement \( s_{AC} = 300 – 100 = 200 \, \text{m} \)
Average velocity \( v_{AC} = \frac{s_{AC}}{t_{AC}} = \frac{200}{210} = 0.95 \, \text{m/s} \)

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic, and the average speed is 30 km/h. What is the average speed for Abdul’s trip?

Solution:

Let the distance one way be \( d \, \text{km} \).
Time for the trip to school \( t_1 = \frac{d}{20} \, \text{hours} \)
Time for the return trip \( t_2 = \frac{d}{30} \, \text{hours} \)
Total distance for the round trip \( D = 2d \, \text{km} \)
Total time for the round trip \( T = t_1 + t_2 = \frac{d}{20} + \frac{d}{30} = \frac{3d + 2d}{60} = \frac{5d}{60} = \frac{d}{12} \, \text{hours} \)
Average speed for the round trip:
\[
v_{\text{avg}} = \frac{D}{T} = \frac{2d}{\frac{d}{12}} = 2 \times 12 = 24 \, \text{km/h}
\]

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 s. How far does the boat travel during this time?

Solution:

Initial velocity \( u = 0 \, \text{m/s} \)
Acceleration \( a = 3.0 \, \text{m/s}^2 \)
Time \( t = 8.0 \, \text{s} \)

Distance travelled:
\[
s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 3.0 \times (8.0)^2 = \frac{3.0 \times 64}{2} = 96 \, \text{m}
\]

5. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. How far does the car travel after applying the brakes?

Solution:

Initial speed:
\[
u = 52 \, \text{km/h} = \frac{52 \times 1000}{3600} \, \text{m/s} = 14.44 \, \text{m/s}
\]
Final speed \( v = 0 \, \text{m/s} \)
Time \( t = 5 \, \text{s} \)

Acceleration:
\[
a = \frac{v – u}{t} = \frac{0 – 14.44}{5} = -2.89 \, \text{m/s}^2
\]

Distance travelled:
\[
s = ut + \frac{1}{2}at^2 = 14.44 \times 5 + \frac{1}{2} \times (-2.89) \times (5)^2
\]
\[
s = 72.2 – 36.125 = 36.075 \, \text{m}
\]

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? After what time will it strike the ground?

Solution:

Final velocity:
Initial velocity \( u = 0 \, \text{m/s} \)
Acceleration \( a = 10 \, \text{m/s}^2 \)
Distance \( s = 20 \, \text{m} \)

Using \( v^2 = u^2 + 2as \):
\[
v^2 = 0 + 2 \times 10 \times 20 = 400
\]
Final velocity:
\[
v = \sqrt{400} = 20 \, \text{m/s}
\]

Time taken:
Using \( v = u + at \):
\[
20 = 0 + 10t
\]
Time:
\[
t = \frac{20}{10} = 2 \, \text{s}
\]

9. State which of the following situations are possible and give an example for each of these:

(a) An object with a constant acceleration but with zero velocity

Solution:
Possible.
Example: A ball thrown vertically upward reaches its highest point and starts to fall back. At the highest point, the velocity is zero, but it has a constant acceleration due to gravity (9.8 m/s²).

(b) An object moving with an acceleration but with uniform speed

Solution:
Not possible.
Explanation: If an object is accelerating, its speed cannot remain uniform. Acceleration implies a change in the velocity, which includes changes in speed or direction.

(c) An object moving in a certain direction with an acceleration in the perpendicular direction

Solution:
Possible.
Example: A car moving in a circular path at a constant speed. The acceleration is directed towards the center of the circle (centripetal acceleration), which is perpendicular to the direction of motion.

10. An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Solution:

Circumference of the orbit:
Radius \( r = 42,250 \, \text{km} \)
Circumference \( C = 2\pi r = 2\pi \times 42,250 = 265,434 \, \text{km} \)

Speed of the satellite:
Time period \( T = 24 \, \text{hours} = 24 \times 3600 \, \text{s} = 86,400 \, \text{s} \)
Speed:
\[
v = \frac{C}{T} = \frac{265,434 \times 1000 \, \text{m}}{86,400 \, \text{s}} = 3,072.6 \, \text{m/s}
\]

Motion class 9 extra numerical questions with answers

Question 1

A car starts from rest and accelerates uniformly at \( 3 \, \text{m/s}^2 \) for 5 seconds. What is its final velocity?

Solution:
Using the formula \( v = u + at \):
\[
v = 0 + (3 \, \text{m/s}^2)(5 \, \text{s}) = 15 \, \text{m/s}
\]
Final velocity: \( 15 \, \text{m/s} \)

Question 2

A cyclist moves with a uniform velocity of \( 10 \, \text{m/s} \) for 10 seconds. How far does the cyclist travel?

Solution:
Using the formula \( s = vt \):
\[
s = (10 \, \text{m/s})(10 \, \text{s}) = 100 \, \text{m}
\]
Distance traveled: \( 100 \, \text{m} \)

Question 3

A bus moving at \( 60 \, \text{km/h} \) comes to rest in 10 seconds. What is its acceleration?

Solution:
Convert speed to \( \text{m/s} \):
\[
60 \, \text{km/h} = 60 \times \frac{1000}{3600} = 16.67 \, \text{m/s}
\]
Using \( a = \frac{v – u}{t} \):
\[
a = \frac{0 – 16.67}{10} = -1.67 \, \text{m/s}^2
\]
Acceleration: \( -1.67 \, \text{m/s}^2 \)

Question 4

A ball is thrown vertically upwards with a velocity of \( 20 \, \text{m/s} \). How high does it go? (Take \( g = 10 \, \text{m/s}^2 \))

Solution:
Using \( v^2 = u^2 + 2as \):
\[
0 = (20)^2 + 2(-10)s
\]
\[
s = \frac{-400}{-20} = 20 \, \text{m}
\]
Maximum height: \( 20 \, \text{m} \)

Question 5

A train starts from rest and travels \( 200 \, \text{m} \) in 10 seconds under uniform acceleration. What is the acceleration?

Solution:
Using \( s = ut + \frac{1}{2}at^2 \):
\[
200 = 0 + \frac{1}{2}a(10)^2
\]
\[
200 = 50a \rightarrow a = 4 \, \text{m/s}^2
\]
Acceleration: \( 4 \, \text{m/s}^2 \)

Question 6

A car traveling at \( 20 \, \text{m/s} \) slows down uniformly to rest in \( 5 \, \text{s} \). Find the distance it travels.

Solution:
Using \( s = ut + \frac{1}{2}at^2 \):
\[
a = \frac{v – u}{t} = \frac{0 – 20}{5} = -4 \, \text{m/s}^2
\]
\[
s = (20)(5) + \frac{1}{2}(-4)(5)^2
\]
\[
s = 100 – 50 = 50 \, \text{m}
\]
Distance traveled: \( 50 \, \text{m} \)

Question 7

A cyclist moving with a speed of \( 6 \, \text{m/s} \) increases his speed to \( 10 \, \text{m/s} \) in \( 2 \, \text{s} \). What is the acceleration?

Solution:
Using \( a = \frac{v – u}{t} \):
\[
a = \frac{10 – 6}{2} = 2 \, \text{m/s}^2
\]
Acceleration: \( 2 \, \text{m/s}^2 \)

Question 8

A body starts from rest and moves with a uniform acceleration of \( 2 \, \text{m/s}^2 \). Find its velocity after traveling \( 16 \, \text{m} \).

Solution:
Using \( v^2 = u^2 + 2as \):
\[
v^2 = 0 + 2(2)(16) = 64 \rightarrow v = 8 \, \text{m/s}
\]
Final velocity: \( 8 \, \text{m/s} \)

Question 9

A car travels around a circular path of radius \( 10 \, \text{m} \) with a uniform speed. If it completes one round in \( 20 \, \text{s} \), find its speed.

Solution:
Using \( v = \frac{2\pi r}{t} \):
\[
v = \frac{2 \times 3.14 \times 10}{20} = 3.14 \, \text{m/s}
\]
Speed: \( 3.14 \, \text{m/s} \)

Question 10

A train moves with a uniform velocity of \( 90 \, \text{km/h} \) for \( 4 \, \text{h} \). How far does the train travel?

Solution:
Using \( s = vt \):
\[
s = (90 \, \text{km/h})(4 \, \text{h}) = 360 \, \text{km}
\]
Distance traveled: \( 360 \, \text{km} \)

Question 11

A car starts from rest and accelerates uniformly at \( 2 \, \text{m/s}^2 \) for \( 8 \, \text{s} \). It then maintains a constant velocity for \( 12 \, \text{s} \) before decelerating uniformly to rest in \( 4 \, \text{s} \). Find the total distance covered by the car.

Solution:
1. Using \( v = u + at \) for the acceleration phase:
\[
v = 0 + (2 \, \text{m/s}^2)(8 \, \text{s}) = 16 \, \text{m/s}
\]
2. Distance during acceleration phase:
\[
s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(8)^2 = 64 \, \text{m}
\]
3. Distance during constant velocity phase:
\[
s_2 = vt = (16)(12) = 192 \, \text{m}
\]
4. Using \( s = vt – \frac{1}{2}at^2 \) for deceleration phase:
\[
a = \frac{v – u}{t} = \frac{0 – 16}{4} = -4 \, \text{m/s}^2
\]
\[
s_3 = (16)(4) + \frac{1}{2}(-4)(4)^2 = 64 – 32 = 32 \, \text{m}
\]
Total distance:
\[
s_1 + s_2 + s_3 = 64 + 192 + 32 = 288 \, \text{m}
\]
Total distance covered: \( 288 \, \text{m} \)

Question 12

A ball is dropped from a height of \( 45 \, \text{m} \). How long does it take to hit the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Solution:
Using \( s = \frac{1}{2}gt^2 \):
\[
45 = \frac{1}{2}(10)t^2
\]
\[
t^2 = 9 \rightarrow t = 3 \, \text{s}
\]
Time to hit the ground: \( 3 \, \text{s} \)

Question 13

A train moving at \( 54 \, \text{km/h} \) slows down uniformly and stops in \( 30 \, \text{s} \). Find the distance it covers before stopping.

Solution:
Convert speed to \( \text{m/s} \):
\[
54 \, \text{km/h} = 54 \times \frac{1000}{3600} = 15 \, \text{m/s}
\]
Using \( a = \frac{v – u}{t} \):
\[
a = \frac{0 – 15}{30} = -0.5 \, \text{m/s}^2
\]
Using \( s = ut + \frac{1}{2}at^2 \):
\[
s = (15)(30) + \frac{1}{2}(-0.5)(30)^2
\]
\[
s = 450 – 225 = 225 \, \text{m}
\]
Distance covered: \( 225 \, \text{m} \)

Question 14

A stone is thrown vertically upwards with an initial velocity of \( 25 \, \text{m/s} \). Find the total time it takes to return to the thrower. (Take \( g = 10 \, \text{m/s}^2 \))

Solution:
Time to reach maximum height:
Using \( v = u + at \):
\[
0 = 25 – 10t \rightarrow t = 2.5 \, \text{s}
\]
Total time for the round trip:
\[
T = 2 \times 2.5 = 5 \, \text{s}
\]
Total time: \( 5 \, \text{s} \)

Question 15

A car accelerates from \( 10 \, \text{m/s} \) to \( 20 \, \text{m/s} \) in \( 4 \, \text{s} \). Find the distance traveled during this time.

Solution:
Using \( s = ut + \frac{1}{2}at^2 \):
\[
a = \frac{v – u}{t} = \frac{20 – 10}{4} = 2.5 \, \text{m/s}^2
\]
\[
s = (10)(4) + \frac{1}{2}(2.5)(4)^2
\]
\[
s = 40 + 20 = 60 \, \text{m}
\]
Distance traveled: \( 60 \, \text{m} \)

Question 16

An artificial satellite revolves around the Earth in a circular orbit of radius \( 42,250 \, \text{km} \) in \( 24 \, \text{hours} \). Find its speed. (Take \( \pi = 3.14 \))

Solution:
Using \( v = \frac{2\pi r}{t} \):
Convert radius and time:
\[
r = 42,250 \times 1000 \, \text{m}, \, t = 24 \times 3600 \, \text{s}
\]
\[
v = \frac{2 \times 3.14 \times 42,250,000}{86,400} = 3,072 \, \text{m/s}
\]
Speed: \( 3,072 \, \text{m/s} \)

Question 17

A vehicle traveling at \( 30 \, \text{m/s} \) decelerates uniformly at \( 3 \, \text{m/s}^2 \). Find the time it takes to stop.

Solution:
Using \( t = \frac{v – u}{a} \):
\[
t = \frac{0 – 30}{-3} = 10 \, \text{s}
\]
Time to stop: \( 10 \, \text{s} \)

Question 18

A particle covers a distance of \( 10 \, \text{m} \) in the first second, \( 20 \, \text{m} \) in the next second, and \( 30 \, \text{m} \) in the third second. Find the acceleration.

Solution:
Total distance in 3 seconds:
\[
s = 10 + 20 + 30 = 60 \, \text{m}
\]
Using \( s = ut + \frac{1}{2}at^2 \):
Here, \( u = 10 \, \text{m/s} \), \( t = 3 \, \text{s} \), \( s = 60 \, \text{m} \):
\[
60 = 10(3) + \frac{1}{2}a(3)^2
\]
\[
60 = 30 + \frac{9}{2}a \rightarrow a = \frac{20}{9} \, \text{m/s}^2
\]
Acceleration: \( 2.22 \, \text{m/s}^2 \)

Question 19

An object travels \( 200 \, \text{m} \) with a uniform speed of \( 10 \, \text{m/s} \) and then \( 100 \, \text{m} \) at \( 5 \, \text{m/s} \). What is its average speed?

Solution:
Total time:
\[
t_1 = \frac{200}{10} = 20 \, \text{s}, \, t_2 = \frac{100}{5} = 20 \, \text{s}
\]
Total distance:
\[
200 + 100 = 300 \, \text{m}
\]
\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}
\]
\[
v_{\text{avg}} = \frac{300}{20 + 20} = \frac{300}{40} = 7.5 \, \text{m/s}
\]
Average speed: \( 7.5 \, \text{m/s} \)

Question 20

A stone is thrown vertically upwards and reaches a height of \( 80 \, \text{m} \). Find its initial velocity. (Take \( g = 10 \, \text{m/s}^2 \))

Solution:
Using \( v^2 = u^2 + 2as \):
\[
0 = u^2 + 2(-10)(80)
\]
\[
u^2 = 1600 \rightarrow u = \sqrt{1600} = 40 \, \text{m/s}
\]
Initial velocity: \( 40 \, \text{m/s} \)

Question 21

A rocket starts from rest and moves upward with a constant acceleration of \( 20 \, \text{m/s}^2 \). After \( 10 \, \text{s} \), its fuel is exhausted, and it continues to move up under gravity. Find the maximum height reached by the rocket. (Take \( g = 10 \, \text{m/s}^2 \))

Solution:
1. Height covered during powered flight:
Using \( s = ut + \frac{1}{2}at^2 \):
\[
s_1 = 0 + \frac{1}{2}(20)(10)^2 = 1000 \, \text{m}
\]
2. Velocity at the end of powered flight:
\[
v = u + at = 0 + (20)(10) = 200 \, \text{m/s}
\]
3. Additional height after fuel exhaustion:
Using \( v^2 = u^2 + 2as \):
\[
0 = (200)^2 + 2(-10)s_2 \rightarrow s_2 = 2000 \, \text{m}
\]
Total height:
\[
s_1 + s_2 = 1000 + 2000 = 3000 \, \text{m}
\]
Maximum height: \( 3000 \, \text{m} \)

Question 22

A car traveling at \( 72 \, \text{km/h} \) overtakes another car moving at \( 54 \, \text{km/h} \). If the length of each car is \( 5 \, \text{m} \), find the time taken by the first car to completely overtake the second car.

Solution:
Convert speeds to \( \text{m/s} \):
\[
v_1 = 72 \times \frac{1000}{3600} = 20 \, \text{m/s}
\]
\[
v_2 = 54 \times \frac{1000}{3600} = 15 \, \text{m/s}
\]
Relative speed:
\[
v = v_1 – v_2 = 20 – 15 = 5 \, \text{m/s}
\]
Total distance to overtake:
\[
5 + 5 = 10 \, \text{m}
\]
Time taken:
\[
t = \frac{s}{v} = \frac{10}{5} = 2 \, \text{s}
\]
Time taken: \( 2 \, \text{s} \)

Question 23

A body falls freely from a height and covers half the total distance in the last second of its fall. Find the total time of its fall.

Solution:
Let the total time be \( T \, \text{seconds} \). Using the formula for distance covered in the last second:
\[
\text{Distance in the last second} = u + \frac{1}{2}g(2T – 1)
\]
Total distance covered:
\[
H = \frac{1}{2}gT^2
\]
Given:
\[
\frac{1}{2}gT^2 = u + \frac{1}{2}g(2T – 1)
\]
After simplifying:
\[
T = 2 \, \text{seconds}
\]
Total time of fall: \( 2 \, \text{seconds} \)

Question 24

An object is thrown upward with a speed of \( 30 \, \text{m/s} \). Find the total time it spends in the air before returning to the ground. (Take \( g = 10 \, \text{m/s}^2 \))

Solution:
1. Time to reach the maximum height:
Using \( v = u + at \):
\[
0 = 30 – 10t_1 \rightarrow t_1 = 3 \, \text{seconds}
\]
2. Total time for upward and downward journey:
\[
T = 2 \times t_1 = 2 \times 3 = 6 \, \text{seconds}
\]
Total time in air: \( 6 \, \text{seconds} \)

Question 25

A bullet is fired horizontally with a speed of \( 100 \, \text{m/s} \) from a height of \( 80 \, \text{m} \). How far will it travel horizontally before hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Solution:
1. Time taken to hit the ground:
Using \( s = \frac{1}{2}gt^2 \):
\[
80 = \frac{1}{2}(10)t^2 \rightarrow t^2 = 16 \rightarrow t = 4 \, \text{seconds}
\]
2. Horizontal distance traveled:
\[
\text{Horizontal distance} = v \times t = 100 \times 4 = 400 \, \text{m}
\]
Horizontal distance: \( 400 \, \text{m} \)

Question 26

A car travels at a constant speed of \( 60 \, \text{km/h} \). How long will it take to cover a distance of \( 180 \, \text{km} \)?

Solution:
Using \( t = \frac{d}{v} \):
\[
t = \frac{180}{60} = 3 \, \text{hours}
\]
Time taken: \( 3 \, \text{hours} \)

Question 27

An object starts from rest and accelerates uniformly at \( 5 \, \text{m/s}^2 \) for \( 10 \, \text{seconds} \). What is the final velocity?

Solution:
Using \( v = u + at \):
\[
v = 0 + 5 \times 10 = 50 \, \text{m/s}
\]
Final velocity: \( 50 \, \text{m/s} \)

Question 28

A cyclist covers a distance of \( 500 \, \text{m} \) in \( 2 \, \text{minutes} \). Calculate the average speed in \( \text{m/s} \).

Solution:
First, convert the time to seconds:
\[
2 \, \text{minutes} = 120 \, \text{seconds}
\]
Using \( v = \frac{d}{t} \):
\[
v = \frac{500}{120} = 4.17 \, \text{m/s}
\]
Average speed: \( 4.17 \, \text{m/s} \)

Question 29

A ball is thrown vertically upward with a velocity of \( 20 \, \text{m/s} \). How high will it go before coming to rest momentarily?

Solution:
Using \( v^2 = u^2 + 2as \):
\[
0 = 20^2 + 2(-9.8)s
\]
\[
0 = 400 – 19.6s \rightarrow 19.6s = 400 \rightarrow s = \frac{400}{19.6} = 20.41 \, \text{m}
\]
Maximum height: \( 20.41 \, \text{m} \)

Question 30

A car accelerates from rest at a constant rate of \( 4 \, \text{m/s}^2 \). How much distance will it cover in \( 5 \, \text{seconds} \)?

Solution:
Using \( s = ut + \frac{1}{2}at^2 \):
\[
s = 0 + \frac{1}{2} \times 4 \times 5^2
\]
\[
s = 2 \times 25 = 50 \, \text{m}
\]
Distance covered: \( 50 \, \text{m} \)

Question 31

A train moving with a uniform velocity of \( 30 \, \text{m/s} \) passes a pole in \( 20 \, \text{seconds} \). What is the length of the train?

Solution:
Using the formula \( d = vt \):
\[
v = 30 \, \text{m/s}, \, t = 20 \, \text{s}
\]
\[
d = 30 \times 20 = 600 \, \text{m}
\]
Length of the train: \( 600 \, \text{m} \)

Question 32

A car decelerates uniformly from \( 25 \, \text{m/s} \) to rest in \( 10 \, \text{seconds} \). Calculate the acceleration and the distance covered.

Solution:
Using the equation \( v = u + at \):
\[
v = 0, \, u = 25 \, \text{m/s}, \, t = 10 \, \text{s}
\]
\[
0 = 25 + 10a \rightarrow 10a = -25 \rightarrow a = -2.5 \, \text{m/s}^2
\]
Now, using \( s = ut + \frac{1}{2}at^2 \):
\[
s = 25 \times 10 + \frac{1}{2}(-2.5)(10)^2
\]
\[
s = 250 – 125 = 125 \, \text{m}
\]
Acceleration: \( -2.5 \, \text{m/s}^2 \)
Distance covered: \( 125 \, \text{m} \)

Question 33

A runner completes one round of a circular track of radius \( 50 \, \text{m} \) in \( 40 \, \text{seconds} \). Calculate the runner’s speed and the distance covered in \( 3 \, \text{minutes} \).

Solution:
The circumference of the track is:
\[
C = 2\pi r = 2\pi \times 50 = 100\pi \, \text{m}
\]
The runner’s speed is:
\[
v = \frac{100\pi}{40} = 7.85 \, \text{m/s}
\]
In \( 3 \, \text{minutes} = 180 \, \text{seconds} \), the distance covered is:
\[
d = v \times t = 7.85 \times 180 = 1413 \, \text{m}
\]
Runner’s speed: \( 7.85 \, \text{m/s} \)
Distance covered: \( 1413 \, \text{m} \)