Motion

PHYSICS

 Question : 1 - An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer:-

Yes, zero displacement is possible if an object has moved through a distance.

Suppose a ball starts moving from point A and it returns back at same point A, then the distance will be equal to 20 meters while displacement will be zero.

Question : 2 - A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:-

Given, side of the square field = 10m

Therefore, perimeter = 10 m x 4 = 40 m

Farmer moves along the boundary in 40s.

Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?

Since in 40 s farmer moves 40 m

Question : 3 - Which of the following is true for displacement?

(a) It cannot be zero.

(b) Its magnitude is greater than the distance travelled by the object.

Answer: None

Question : 4 - Distinguish between speed and velocity.

Answer:

Speed has only magnitude while velocity has both magnitude and direction.

Question : 5 - Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer: The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity.

Question : 6 - What does the odometer of an automobile measure?

Answer: In automobiles, odometer is used to measure the distance.

Question : 7 - What does the path of an object look like when it is in uniform motion?

Answer: In the case of uniform motion the path of an object will look like a straight line.

 

Solution of In Text Questions - 2

Question :- 8 - During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ ms^-1.

Answer:-

Question : 9 - When will you say a body is in

(i) uniform acceleration?

(ii) non-uniform acceleration?

Answer:

(i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.

(ii)A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

Question : 10 - A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Answer:

Question : 11 - A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

Answer:

Here we have,

Initial velocity, u = 0,

Final velocity, v = 40km/h = 11.11m/s

Time (t) = 10 minute = 60x10=600s

Acceleration (a) =?

Question : 12 - What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer:

(a) The slope of the distance-time graph for an object in uniform motion is straight line.

(b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

Question : 13 - What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer:

When the slope of distance-time graph is a straight line parallel to time axis, the object is moving with uniform motion.

Question : 14 - What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer:

When the slope of a speed time graph is a straight line parallel to the time axis, the object is moving with uniform speed.

Solution of In Text Questions - 3

Question : 15 - What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer:-

The quantity of distance is measured by the area occupied below the velocity time graph.

Question :- 16 - A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled.

Answer:

Here we have,

Initial velocity (u) = 0

Acceleration (a) = 0.1ms-2

Time (t) = 2 minute = 120 second

(a) The speed acquired:

We know that, v = u + at

⇒ v = 0 + 0.1m/s2 x 120 s

⇒ v = 120 m/s

Thus, the bus will acquire a speed of 120 m/s after 2 minute with the given acceleration.

(b) The distance travelled:

Question : 17 - A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s^-2. Find how far the train will go before it is brought to rest.

Answer:

Here,we have,

Question : 18 - A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?

Answer:

Here we have,

Initial velocity, u = 0

Acceleration (a) = 2cm/s² = 0.02m/s²

Time (t) = 3s

Therefore, Final velocity, v =?

Question : 19 - A racing car has a uniform acceleration of 4 m s^-2. What distance will it cover in 10 s after start?

Answer:

Here we have,

Acceleration, a = 4m/s2

Initial velocity, u =0

Time, t = 10s

Therefore, Distance (s) covered =?

Thus, racing car will cover a distance of 200m after start in 10 s with given acceleration.

Question : 20 - A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer:

Here we have,

Initial velocity (u) = 5m/s

Final velocity (v) =0 (Since from where stone starts falling its velocity will become zero)

Acceleration (a) = -10m/s²

(Since given acceleration is in downward direction, i.e. the velocity of the stone is decreasing, thus acceleration is taken as negative)

Height, i.e. Distance, s =?

Time (t) taken to reach the height =?

Solution of NCERT Exercise (Part 1)

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:-

Here we have,

Diameter = 200 m, therefore, radius = 200m/2 = 100 m

Time of one rotation = 40s

Time after 2m20s = 2 x 60s + 20s = 140s

Distance after 140 s = ?

Displacement after 140s =?

 

(a) Distance after 140s

We know that,distance=velocity ×time

⇒distance=15.7m/s ×140 s = 2198 m

(b) Displacement after 2 m 20 s i.e. in 140 s

Since,rotatin in 40 s=1

Therefore, in 3.5 rotations athlete will be just at the opposite side of the circular track, i.e. at a distance equal to the diameter of the circular track which is equal to 200 m.

Therefore,

Distance covered in 2 m 20 s = 2198 m

And, displacement after 2 m 20 s = 200m

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:-

Here we have,

Distance from point A to B = 300 m

Time taken = 2 minute 30 second = 2 x 60 + 30 s = 150 s

Distance from point B to C = 100 m

Time taken = 1 minute = 60 s

(a) Average speed and velocity from point A to B

(b) Average speed and velocity from B to C

Therefore,average velocity=1.66 m/s west

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:

Strategy: We need to calculate the time taken in each of the trip. After that, we can calculate the average speed.

Let the distance of the school = s km

Let time to reach the school in first trip = t₁

Let time to reach the school in second trip = t₂

Therefore, average speed of Adbul = 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?

Answer:

Here we have,

Initial velocity (u) = 0

Acceleration (a) = 3.0m/s²

Time = 8 s

Therefore, distance (s) covered =?

Therefore, boat travel a distance of 96 m in the given time.

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:

Given for first driver,

 

In the graph, blue slope shows the velocity of the first car and green slope shows the velocity of the second car.

Distance is calculated by the area under the slope of the graph.

Solution of NCERT Exercise (Part 2)

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) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.

(b) All of them never come at the same point at the same time.

(c) According to graph; each small division shows about 0.57 km.

A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km

Thus, at this point C travels about

9.14 – (0.57 x 3.75) km = 9.14 km – 2.1375 km = 7.0025 km ≈ 7 km

Thus, when A passes B, C travels about 7 km.

(d) B passes C at point Q at the distance axis which is ≈ 4km + 0.57km x 2.25 = 5.28 km

Therefore, B travelled about 5.28 km when passes to C.

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?

Answer:-

Here we have,

Initial velocity,u=0

Distance,s=20m

Acceleration,a= 10 m s^-2

Final velocity,v=?

Time,t= ?

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) Distance travelled by car in the 4 second

The area under the slope of the speed – time graph gives the distance travelled by an object.

In the given graph

56 full squares and 12 half squares come under the area slope for the time of 4 second.

Total number of squares = 56 + 12/2 = 62 squares

The total area of the squares will give the distance travelled by the car in 4 second.

(b) Part MN of the slope of the graph is straight line parallel to the time axis, thus this potion of graph represents uniform motion of 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:

(a) The term acceleration implies that the velocity of the object is changing; inspite of that constant acceleration with zero velocity is impossible. When an object is thrown in upward direction, at the maximum height the velocity of the object becomes zero but still in that condition a constant acceleration due to gravity is working.

(b) Object moving in a certain direction with an acceleration in perpendicular direction is possible; in case of circular motion. When an object moves on a circular path, its direction is along the tangent of the circle but acceleration is towards the radius of the circle. We know, that a tangent always makes a right angle with the radius; so when an object is in circular motion, the acceleration and velocity are in mutually perpendicular direction.

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:

Here we have,

Radius, r = 42250km

Time, t = 24 hours

Speed =?