In this post we are going to show the detailed solution for class 9th science chapter 8. All the solution are prepared by our esteemed who are very well experienced in the teaching.
Chapter 8 Motion
Question 1. An athlete completes one round of a circular track of diameter 200m in 40 seconds. What will be the distance covered and the displacement at the end of 2 minutes 20 seconds?
Solution : Diameter of a track is : 200m
therefore, radius of the track will be : 200/2 = 100m
Time taken to complete 1 round = 40s
total times the athlete moves = 2 minutes 20 seconds
= 2 × 60 + 20 seconds
= 140 seconds
Distance covered in one round = Circumference of a circle
= 2πr
= 2 × π × 100
= 200 π m
Thus,
Distance covered in 40 seconds = 200 π
Distance covered in 1 second = 200 π / 40
= 5 π meter
Now,
Distance covered in 40 second = 140 × 5 π
= 700 π
= 700 × 22/7
= 100 × 22
= 2200m
Finding displacement,
Number of rounds completed by the athlete in 140 seconds = 140/ 40
= 3.5 rounds.
Therefore, the final position of the athlete is at the opposite end of the circular track . Therefore, the net displacement will be equal to the diameter of the track which is 200 m .
The net distance covered by the athlete is 22000 meter and the displacement of the athlete is 200m.
Therefore the distance is 2200 meter, and the displacement is 200m.
Question 2. Joseph jogs from one end A to the other end B of a straight 300m road in 2 minutes 30 seconds and then turn around and jogs 100m back to point C in another 1 minute. What are joseph’s average speed and velocities in jogging (a) from A to B and (b) from A to C ?
Solution : Given :
Distance covered from point A to point B = 300m
Distance covered from point A to point C = 300 + 100 = 400m
Time taken to travel from point A to point B = 2 minute 30 second = 2 × 60 + 30 = 150 seconds
Time taken to travel from point A to point C = 2 minute 30 second +1 minute = 210 seconds
Displacement from A to B = 300m
Displacement from A to C = 300 – 100 = 200m
Average speed = total distance travelled /total time taken
Average velocity = total displacement / time taken
Therefore, the average speed while travelling from A to B = 300/150ms¯¹ = 2 m/s
Average speed while travelling from A to C = 400/210 ms¯¹ = 1.9 m/s
Average velocity while travelling from A to B = 300/150 ms¯¹ = 2 m/s ( Direction A to B)
Average velocity while travelling from A to C = 200/210 ms¯¹ = 0.95 m/s.(Direction A to C)
Question 3. Abdul, While driving to school, computers the average speed for his trip to be 20km h¯¹. On his return trip along the same route , there is less traffic and the average speed is 30 km h¯¹. What is the average speed for Abdul’s trip ?
Solution : Distance travelled to reach the school = Distance travelled to reach the home (d)
Time taken to reach to school = t1
Time taken to reach to home = t2
Therefore, average speed while going to school = total distance travelled / total time taken = d/ t1 = 20km/h
Average speed while going to home = total distance travelled / total time taken = d/ t2 = 30 km/h
Therefore, t1 = d/20 and t2 = d/30
Now, the average speed for the entire trip is given by total distance travelled / total time taken
= (d+d) / (t1+t2) km/h = (d+d) / (d/20+d/30) km/h
= 120/5 kmh¯¹ = 24 kmh¯¹
Therefore, Abdul’s average speed for the entire trip is 24 kilometer per hour.
Question 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms ¯² for 8.0 s. How far does the boat travel during this time?
Solution : Since motorboat start from rest,
Its initial velocity will be u = 0
Therefore,
Initial velocity of the boat – u = 0
Acceleration of the boat – a = 3 m/s²
Time period – t = 8.0 s
We need to find how far the boat will go = Distance
Since we know u , a , and t.
As per the second equation of motion to find distance (s)
s = ut + ½ at²
s = 0 × 8+ ½ × 3 × 8²
s = 0 + ½ × 3 × 64
s = 3÷2 × 64
s = 3 × 32
s = 96 m.
Therefore, the motorboat travels a distance of 96 m in the time of 8 seconds.
Question 5. A driver of a car travelling at 52 km h‾¹ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h‾¹ in another car applies his brakes slowly and stop in 10s. On the same graph paper, plot the speed versus time graph for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution :
First let us convert both into m/s
Car A :
Initial velocity of car a = 52 km/h
= 52 × 5/18 m/s
= 130/ 18 m/s
= 14.44 m/s
Final velocity = 0 m/s
Time = 5 seconds
Car B :
Initial velocity of Car B = 3 km/h
= 3 × 5/ 18 m/s
= 5/6 m/s
= 0.833 m/s
Final velocity = 0 m/s
Time = 10 seconds
Car A :
Distance travelled = Area under velocity time graph
= ½ × b × h
= ½ × 5 × 130 /9
= 36.11 m
Car B :
Distance travelled = Area under velocity time graph
= ½ × b × h
= ½ × 10 × 5/ 6
= 4.166 m
Therefore, the first Car A (which was travelling at 52 km/h) travelled farther when brakes were applied.
Question 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) Which of the three is travelling the fastest ?
(b) Are all three ever at the same point on the road ?
(c) How far has C travelled when B passes A ?
(d) How far has B travelled by the time it passes C ?
Answer –
(a) Speed = Distance / Time
From the graph we can see that,
A travels from 6 to 12 km in 2 hours
B travels from 0 to 12 km in 1.4 hours
C travels from 2 to 12 km in 1.6 hours
Since, B travels the most distance in small period of time
Therefore, B is fastest,
(b) Since the three lines do not intersect at a same point, the three objects never meet at the same point on the road.
(c) Since there are 7 unit area of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.
since the initial point of object, C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7) km = 16/7 km.
When A passes B, the distance between the origin and C is 8 km
Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km
(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units.
Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km.
Question 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 ?
Solution – Distance travelled – s = 20m
Acceleration = 10 m/s²
Since, the ball is dropped
Its initial velocity will be o
Its initial velocity = u = 0 m/s
We need to find with velocity will it strike the ground
i.e. = final velocity
We know u, a and s finding v using third equation of motion
⇒ v² – u² = 2as
⇒ v² – (0)² = 2 × 10 × 20
⇒ v² = 400
⇒ v = 400 (square root)
⇒ v = 20 m/s
Since, acceleration is positive,
it means that velocity will increase
So, velocity will be positive
⇒ v = 20 m/s
Now, we need to find after what time will it strike the ground
By using the first equation of motion
⇒ v = u + at
⇒ 20 = 0 + 10 t
⇒ 10 t = 20
⇒ t = 20 /10
⇒ t = 2 second
Therefore, the ball reaches the ground after 2 second.
Question 8. The speed – time graph for a car is shown is Fig 8. 12.
Fig 8.12
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represent the distance travelled by the car during the period.
Answer – 
The shaded area represents the displacement of the car over a time period of 4 seconds. it can be calculated as :
(½) × 4 × 6 = 12 meters.
Therefore, the car travels a total of 12 meters in the first four seconds.
(b) Which part of the graph represents uniform motion of the car ?
Answer – Since the speed of the car does not change from the points (x= 6) and (x=10), the car is said to be in uniform motion from the 6th to the 10th second.
Question 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
Answer – It is possible an object thrown up into the air has a constant acceleration due to gravity acting on it . However, when it reaches its maximum height , its velocity is zero.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer – It is possible ; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time.
Question 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 – Given, radius of the orbit = 42250 km
Therefore, circumference of the orbit = 2 × π × 42250 km = 265571.42 km
Time taken for the orbit = 24 hours
Therefore, speed of the satellite = 11065.4 km.h-1
The satellite orbits the Earth at a speed of 11065.4 kilometer per hour.
You can see other posts here:
- NCERT Class 9th Science Chapter 8 – Motion Summary
- NCERT Science Class 9th Chapter 9 Force and Laws of Motion Summary
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