Formulas Used in solving the Numerical Problems
Average Speed:
Average speed = Total distance travelled / Total time taken
If an object travels a distance s in time t, then its speed v is given by:
v = s / t
Average Velocity:
Average velocity = (Initial velocity + Final velocity) / 2
Mathematically:
v_av = (u + v) / 2
Equations of Motion:
Equation 1:
v = u + at
Equation 2:
s = ut + 1/2 x a x t²
Equation 3:
2as = v² – u²
Uniform Circular Motion:
When an object moves in a circular path with uniform speed, its speed v is given by:
v = (2 x π x r) / t
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 = 2πr = 2π × 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 dAB = 300 m
Time tAB = 2 minutes 30 seconds = 150 s
Average speed vavg, AB = dAB/tAB = 300/150 = 2 m/s
Since he jogs in a straight line from A to B, the average velocity vAB = vavg, AB = 2 m/s
From A to C:
Distance dAC = 300 + 100 = 400 m
Time tAC = 150 + 60 = 210 s
Average speed vavg, AC = dAC/tAC = 400/210 = 1.9 m/s
Displacement sAC = 300 – 100 = 200 m (as C is 100 m back from B)
Average velocity vAC = sAC/tAC = 200/210 = 0.95 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 km.
Time for the trip to school, t1 = d/20 hours
Time for the return trip, t2 = d/30 hours
Total distance for the round trip, D = 2d km
Total time for the round trip, T = t1 + t2 = d/20 + d/30 = (3d + 2d)/60 = 5d/60 = d/12 hours
Average speed for the round trip, vavg = D/T = 2d/(d/12) = 2 × 12 = 24 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 m/s
Acceleration a = 3.0 m/s²
Time t = 8.0 s
Distance travelled, s = ut + 1/2at² = 0 + 1/2 × 3.0 × 8.0² = (3.0 × 64)/2 = 96 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 km/h = (52 × 1000)/3600 m/s = 14.44 m/s
Final speed v = 0 m/s
Time t = 5 s
Acceleration a = (v – u)/t = (0 – 14.44)/5 = -2.89 m/s²
Distance travelled, s = ut + 1/2at² = 14.44 × 5 + 1/2 × -2.89 × 5² = 72.2 – 36.125 = 36.075 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 m/s
Acceleration a = 10 m/s²
Distance s = 20 m
Using v² = u² + 2as, v² = 0 + 2 × 10 × 20 = 400
Final velocity v = √400 = 20 m/s
Time taken:
Using v = u + at, 20 = 0 + 10t
Time t = 20/10 = 2 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 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
Circumference of the orbit:
Radius r = 42250 km
Circumference C = 2πr = 2π × 42250 = 265,434 km
Speed of the satellite:
Time period T = 24 hours = 24 × 3600 s = 86400 s
Speed v = C/T = (265,434 × 1000 m)/86400 s = 3072.6 m/s
Motion class 9 extra numerical questions with answers
1. A car travels at a constant speed of 60 km/h. How long will it take to cover a distance of 180 km?
Solution:
Use the formula t = d / v.
Here, d = 180 km and v = 60 km/h.
So, t = 180 / 60 = 3 hours.
2. An object starts from rest and accelerates uniformly at 5 m/s² for 10 seconds. What is the final velocity?
Solution:
Use the equation v = u + at.
Initial velocity, u = 0 m/s.
Acceleration, a = 5 m/s².
Time, t = 10 s.
So, v = 0 + 5 x 10 = 50 m/s.
3. A cyclist covers a distance of 500 m in 2 minutes. Calculate the average speed in m/s.
Solution:
First, convert the time to seconds: 2 minutes = 120 seconds.
Use the formula v = d / t.
Distance, d = 500 m.
Time, t = 120 s.
So, v = 500 / 120 = 4.17 m/s.
4. A ball is thrown vertically upward with a velocity of 20 m/s. How high will it go before coming to rest momentarily?
Solution:
Use the equation v² = u² + 2as.
Final velocity, v = 0 m/s.
Initial velocity, u = 20 m/s.
Acceleration, a = -9.8 m/s².
0 = 20² + 2 x -9.8 x s.
0 = 400 – 19.6s.
19.6s = 400.
s = 400 / 19.6 = 20.41 m.
5. A car accelerates from rest at a constant rate of 4 m/s². How much distance will it cover in 5 seconds?
Solution:
Use the equation s = ut + 1/2 x at².
Initial velocity, u = 0 m/s.
Acceleration, a = 4 m/s².
Time, t = 5 s.
So, s = 0 + 1/2 x 4 x 5².
s = 2 x 25 = 50 m.
6. A train moving with a uniform velocity of 30 m/s passes a pole in 20 seconds. What is the length of the train?
Solution:
Use the formula d = vt.
Velocity, v = 30 m/s.
Time, t = 20 s.
So, d = 30 x 20 = 600 m.
7. A car decelerates uniformly from 25 m/s to rest in 10 seconds. Calculate the acceleration and the distance covered.
Solution:
Use the equation v = u + at.
Final velocity, v = 0 m/s.
Initial velocity, u = 25 m/s.
Time, t = 10 s.
0 = 25 + 10a.
10a = -25.
a = -2.5 m/s².
Use the equation s = ut + 1/2 x at².
s = 25 x 10 + 1/2 x -2.5 x 10².
s = 250 – 125 = 125 m.
8. A runner completes one round of a circular track of radius 50 m in 40 seconds. Calculate the runner’s speed and the distance covered in 3 minutes.
Solution:
The circumference of the track is C = 2πr = 2π x 50 = 100π m.
The runner’s speed is v = 100π / 40 = 7.85 m/s.
In 3 minutes (180 seconds), the distance covered is d = v x t = 7.85 x 180 = 1413 m.
