NCERT Solutions for Class 9 Science Chapter 8 Motion

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?

Answer: Given that diameter of circular track d = 200m

Radius of track r= d/2 = 100m

We know that circumference = 2 π r

∴ circumference = 2 π (100)= 200 π

In 40 s distance covered by the athlete is 200 m.

∴ In 1 s distance covered by athlete = 200 π / 40

The athlete runs for 2 mins 20 sec

∴ Total distance covered in 140 s = (200 x 22 x 140) / 40 x 7 = 2198 m

In 40 s athlete covers one round. Therefore in athlete will cover 3 rounds in 120 secs, and he must be taking fourth round. Also after completing 3 round he reaches at same starting position therefore total displacement will be zero. Therefore, total displacement will be in 20 s during the fourth round. As displacement is shortest distance from initial and final position, ∴ shortest distance will be diameter of track which is equal to 200 m.

Therefore, distance covered by an athlete in 2 min 20s is 2198 m and displacement 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?

Answer: (a) From A to B –

While jogging Joseph covers distance = 300m

Time taken to cover distance of 300 m = 2 min 50 sec = 170 sec

Average speed = total distance covered / time taken = 300 / 170 = 1.77 m/s

Now average velocity = displacement / time interval = 300 / 170 = 1.77 m/s

∴ The average speed and average velocity of Joseph from A to B is same i.e. 1.77 m/s

(b) From A to C –

Here, average speed = total distance covered / total time taken

∴ total distance covered = distance from A to B + distance from B to C = 300 + 100 = 400 m

Also, total time covered = Time taken to move from A to B + time taken to move from B to C

= 170 + 60 = 230 s

Therefore, average speed = 400 / 230 = 1.74 m/s

Also, we know velocity = displacement / time interval

∴ Displacement from A to C = A to B – B to C = 300 – 100 = 200

Average velocity = 200 / 230 = 0.87 m/s

∴ Average speed is 1.74 m/s and average velocity is 0.87 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?

Answer: Given : Average speed of Abdul’s trip = 20 km/h

We know that, Average speed = total distance / time taken

Let d be the total distance travelled to reach the school and t1 be the total time taken to reach the school.

∴ 20 = d / t1

∴ t1 = d/20

While returning from school distance travelled will be same i.e. d and total time required will be t2

∴ 40 = d / t2

∴ t2 = d / 40

Therefore, average speed of Abdul’s trip = total distance covered / total time taken

= ( d + d ) / (t1 + t2)

∴ average speed = 2d / t1+t2

= 2d / [(d/20) + (d/40)]

= 2d / [3d / 40]

= 80 / 3

= 26.66 m/s

Therefore, average speed of Abdul’s trip is 26.66 m/s.

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?

Answer: As boat is initially at rest therefore its initial velocity is 0 m/s.

Acceleration a of motorboat = 3m/s²

Time required t = 8 s

From second equation of motion,

s = ut + ½ at²

s = 0x8 + ½ x3x8²

s= ½ x3x64

s = 3 x 32

s = 96 m

Distance travelled by boat is 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. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer: Consider Case I –

Initial speed of car U1 = 52 km/h

∴ Initial speed of car in m/s = 14.4 m/s

Time required to stop the car t1 = 5s

Final speed of car becomes zero when brakes are applied for 5 s

Consider Case II –

Initial speed of car U2 = 3 km/h = 0.833m/s ≅ 0.83 m/s

Time taken to stop the car t1 = 10s

Final speed of car becomes zero after 10 s of application of brakes

Speed time graph of two cars can be shown as follows-

Distance travelled by each car can be shown by shaded region in the graph,

Thus, distance covered in case I =

s1 = ½ x OP x OR = ½ x 14.4 x 5 = 36 m

Also, distance covered in case II =

s2 = ½ x OS x OQ = ½ x 0.83 x 10 = 4.15 m

Here area of triangle OPR is greater than area of triangle OSQ

Therefore, distance covered in case I is greater case II, thus, car with speed 52 km / h travels larger distance after brakes were applied.

6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

(a)speed = distance / time, also slope of graph = distance / time.

Therefore, speed = slope of graph

Slope of object B is larger than A and C therefore, B is traveling fastest

(b) Are all three ever at the same point on the road?

Answer: Object A, B and C never meet on single point, this suggest that all three are not on the same point on the road.

(c) How far has C travelled when B passes A?

Answer: On distance axis – 7 boxes are at 4 km

∴ 1 box = 4 / 7 km

At initial stage, C is 4 boxes away from origin

Thus, initial distance of C from origin = 16 / 7 km

Distance of object C from origin, when B passes to A is 8 km

Therefore, distance covered by C = 8 – (16/7) = ( 56 – 16 ) / 7 = 40 / 7 = 5.714 km

(d) How far has B travelled by the time it passes C?

Answer: When D passes to C distance covered is 9 boxes

∴ ( 4 x 9 ) / 7 = 36 / 7 = 5.142 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?

Answer: Given : Distance covered by ball is 20 m

Acceleration a is 10 m/s²

As ball is initially at rest therefore initial velocity is 0.

Let v be the final velocity of the ball striking at the ground.

By third equation of motion,

V² = u² + 2as

V² = 0 + 2 x 10 x 20

V² = 400 m/s

V = 20 m/s

By first equation of motion i.e v = u + at

∴ time t taken by ball to strike on ground 20 = 0 + 10(t)

∴ t = 2 s

∴ ball strikes at ground in 2 sec with velocity of 20 m/s

8. The speed-time graph for a car is shown is Fig. 8.12.

(a) Find how far does the car travel in the first 4 seconds.

Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Answer: (a) Shaded region in graph is equal to ½ x 4 x 6 = 12 m. it represents distance travelled by car in first 4 seconds.

(b) Highlighted part in the graph between 6 to 10 s shows uniform motion of the car.

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

(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer: Both the situations are possible

(a) When ball is thrown at maximum height, its velocity is zero. However its acceleration is constant, which is equal to 9.8 m/s².

(b) Acceleration is perpendicular to its direction when car moves in circular track.

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.

Answer: We know that Speed = distance / time

Distance of circular orbit = 2 π r = 2 x 3.14 x 42250 = 265330

Time is 24 hours

Therefore, speed = distance / time = 265330 / 24 = 11,055. 41 km/ h

Check other lessons NCERT Solutions for Class 9

Download free NCERT textbook class 9 Science – Chapter 8 Motion