ADDITIONAL QUESTIONS AND ANSWERS:
NUMERICAL EXAMPLES:
1. A person moves 50 m from P to Q towards south and then moves 120 m from Q to R towards east. Calculate the distance travelled by the person.
Ans: The distance travelled by the person is 50+ 120 = 170 metres
B. Calculate his displacement.
Ans: The directions south and east are perpendicular to each other. Hence PQ ⊥ QR.Hence PQR is a right angled triangle with PR as hypotenuse which corresponds to the displacement.
Therefore PR 2= PO 2 + QR 2 = 502+ 1202 = 2500+ 14400 = 16900 Therefore PR = 130 m
2. A car starts from rest and attains a velocity of 90 km/h. Calculate
(A). Its acceleration
(B). The distance travelled by the car in 10 s.
Ans: Given u = 0 km /h since it starts from rest,
V = 90 km/ h = 25 m/s and t = 10 s. Find ‘a’.
(A) a = v−u/t = 25−0/10 = 25/10 = 2.5 ms-2
(B).The distance travelled by the car in 10 s.
S = ut +1/2at2 ⇒ 0 x 10 + 1/2 x 2.5 x 102
⇒ 0 + 12.5 = 12.5 m.
3. A passenger bus is accelerated uniformly from 18 kmh-1 to 54 km h-1 in 5s. Find
(A) It’s acceleration. (B) The distance travelled
Ans: Given u = 18 kmh-1=5 m s-1, v = 54 km h-1 = 15 m s-1 and t = 55
∴ Find ‘a’
(A) v − u/t = 15 − 5/5 = 10/5 = 2ms-2 = 2 m/s2.
(B) The distance travelled by the bus in 5 s.
s = ut + 1/2at 2 = 5 x 8 + 1/2 x 2 x 25 = 40 + 25 = 65 m.
4. A train, approaching a station with a velocity of 108 km h-1, applies brakes and comes to a halt after 2 minutes. Find.
(A). The acceleration of the train after applying the brakes.
(B). The distance travelled by the train in 2 m.
Ans: Given u = 108 kmh-1 = 30 ms-1,
v= 0 km h-1 = 0 ms-1 and t = 120 s Find ‘a
(A) a = v−u /t = 0−30/120 = −30/120 = −0.25 ms-1
(B) The distance travelled by the train in 2 m; s = ut+ 1/2at 2
= 30 x 120 +1/2 x (−0.25) x 120 x 120
= 3600−1800 = 1800 m
5. A body dropped from the top of a tower strikes the ground after 4s. Calculate
(A). The height of the tower. (acceleration due to gravity 9.8 ms-2 )
(B). The velocity with which it will strike the ground
Ans: Given u = 0 ms-1, t = 4 s and a/g = 9.8 ms-2 Find height s.
(A). s = ut + 1/2at 2 = 0 x 4 + x 1/0 x 9.8 x 16 = 78.4 m
(B). The velocity with which it strikes the ground.
v = u + at =0 + 9.8 x 4 = 39.2 ms-1
6. A football kicked vertically upwards reaches the highest point in 3 seconds. Calculate
(A). The velocity with which it was kicked up. (Acceleration due to gravity of the earth as 9.8 ms2.
(B).The height reached by the football
Ans: (A) Given final velocity v = 0 ms-1, t = 3 s and a= g = −9.8 m/s2.
Since ‘g’ acts opposite to the direction of motion. Find initial velocity.
V = u + at hence u = v − at = 0 − (−9.8) x 3= 29.4 m s-1.
(B). The height reached by the footballs = ut + 1/2at 2
= 29.4 x 3 + ½ x (−9.8) x 9
88.2−44.1 = 44.1 m.
7. The following figure shows the velocity-time graph of an object.
Study the graph and answer the following questions.
A).State the velocity of the object at the end of 10s.
Ans: 18 ms-1.
B). State the velocity of the object at the end of 20s
Ans: 36 ms-1
C).What is the acceleration of the object
Ans: The velocity of the object changes from 18 ms-1 to 36 ms-1 in 10 seconds
Therefore its acceleration is (361−8)/10 = 18/10 = 1.8ms-2
8. In the following figure OA is a straight line path. B and C are intermediate points answer the questions given below with reference to the figure.
(A). If an object starts from B moves to C and then from C to A
(i) What is the displacement of the object. (ii) How much distance has the object travelled.
(B). If the object travels from O to A and then A to C with a uniform speed of
19 kmh-1 Calculate the time taken to travel from O to A
Ans: A (i) 25 km. (ii) 45 km. B: 5 hours
9. Due to uniform acceleration, the velocity of a body changes from 10 cm/s’ to 50 cm /s’ in 8 s. Calculate the acceleration
Ans: Given u = 10 cm , v = 50 cm s and t = 8 s, find ‘a’
a = v−u /t = 50−10/8 = 40/8 = 5cm s-2
10. ABC is a triangular traffic island with each side 20 m in length. A cyclist riding his cycle at uniform speed starts from A and moves along the sides of the traffic island. If he takes one minute to complete one round, find:
(A) The speed of the cyclist
(B) The distance travelled by the cyclist and his displacement at the end of 5 minutes.
(C) The displacement of the cyclist at the end of (i) 70 seconds.
Ans: (A)The perimeter of the traffic island is 60m. The cyclist covers a distance of 30m in 1 minute i.e. 60 seconds Hence the speed of the cyclist is = 60/60 = 1m/s
(B).The distance travelled = speed x time = 1 x 300 = 300 m. At the end of every one minute the cyclist travels 60m and is at the initial position A. At the end of 5 minutes the cyclist has travelled 300 m. Therefore he has completed (300/60) exactly 5 rounds and is again at the initial position. Therefore his displacement is zero.
(C). (i) The distance travelled in 70 seconds = 70 x 1 = 70 m. Thus after covering 60 m he is at the starting position (displacement = 0). Therefore the cyclist is at a distance of 70 − 60= 10 m from the starting position. Therefore the displacement is 10 m.
11. A body starts from rest and is uniformly accelerated in a straight line at the rate of 5 m/s-1. Calculate
(A). The velocity of the body the end of 10 s.
(B) The distance travelled by the body in the first 6 s.
(C) The distance travelled by the body in the last 4 s.
Ans:
(A) In this case u = 0 since the body starts from rest, a = 5 m/s, t = 10s.
V = u + at = 0+5 x 10 50 m/s.
(B) The distance travelled by the body in the first 6 s.
S = ut −1/2 at2
= 0 x t +1/2 x 5 x 6 2 = 1/2 x 5 x 36 = 90 m
(C): Distance travelled in last 1 second = S6= ½ x a x t62
½ x 5 x 36 = 90 m
Distance travelled in 10 seconds = S10=½ x a x t102
½ x 5 x 100 = 250 m
Distance travelled in last 4 s = 250 − 90 = 160 m.
Answer the following questions:
1. The equation of motion which gives the relationship between.
a) u, v, a and t
Ans: v = u + at or a = v−u/t
b) s, u, a and t
Ans: s = ut +1/2 at 2
c) u, v, a and s
Ans: v 2 − u 2 = 2as
2. Give some examples of circular motion.
Ans: Some examples of circular motion are motion of the moon around the earth, merry-go-round, the outer edge of the blade of a rotating fan.
3. Define the following :
(a) Displacement (b) Speed (c) Velocity (d) Acceleration
Ans: a) Displacement is the shortest distance measured from the initial position to the final position of an object.
b) Speed is the distance travelled in unit time.
c) Velocity is the speed of an object moving in a certain direction.
d) Acceleration is the rate of change of velocity
4. The body travels a certain distance but its displacement is zero.
Illustrate with an example.
Ans: A body moves a distance ‘x metres from point P to Q and back from P. Thus the distance travelled by the body is ‘2 x. But the displacement is zero because the body is at the same initial position.
5. When is a body said to be in uniform motion.
Ans: A body is said to be in uniform motion if it covers equal distances in equal intervals of time.
6. What is average velocity.
Ans: Average velocity is the arithmetic mean of initial and final velocity
7. The body travelling with uniform speed has variable velocity. Give an example of such a situation.
Ans: A body moving in a circular path with uniform speed is an example a body travelling with uniform speed but having a variable velocity.
8. How does velocity differ from speed.
Ans: Speed has no direction but velocity has direction.
Answer the following briefly :
1. When is acceleration said to be (i) positive (ii) negative.
Ans: (i) Acceleration is positive if it is in the direction of motion and
(ii) It is negative if it is opposite to the direction of motion
2.What is the effect of (i) positive acceleration and (ii) negative
Ans: If the acceleration is positive, If the acceleration is negative then the velocity decreases by equal amounts in equal intervals of time.
3. When is the acceleration of a body said to be uniform.
Ans: The velocity constantly increases and the acceleration is said to be uniform if the velocity increases.
4. Graphs can be used to describe motion of an object. Why.
Ans: Graphs show dependence of one physical quantity such as distance or velocity on another quantity such as time.
5. When can the speed of an object be equal to its velocity.
Ans: Speed of an object can be equal to its velocity when the object moves inthe same direction in a straight line.
6. What is the acceleration of a body which travels with uniform velocity.
Ans: The acceleration is zero because there is no change in velocity since the velocity is uniform.
7. Why is it difficult to describe the position of an object in a desert.
Ans: The position of an object can be described only by specifying the reference points. In a desert there is sand, sand and sand everywhere.Therefore the area that surrounds the object is uniform and it is difficult to get a reference point.
8. In a race an athlete may be asked to start at A and end at B or vice versa. Why.
Ans: In a race the physical quantity put to the test is speed. Speed has only magnitude and no direction. Therefore it is immaterial whether the athlete runs from A to B or B to A.
9. Though an athlete runs along a circular track with uniform speed, his velocity is not uniform. Why.
Ans: When an athlete runs along a circular track, the direction of his motion changes at every point of the motion i.e. it is tangential to the circular path. Velocity has magnitude as well direction. Since direction changes constantly, velocity changes at every point.
10. When a stone tied to a string is whirled round and released tangential to its circular path.
Ans: Whenever a body is in circular motion, the direction of its motion is tangential to the circular path.Therefore a stone which is tied to a string, whirled round and released, moves in direction tangential to the circular path of the stone.
11. Though the moon revolves round the earth with a uniform speed still its velocity is not uniform but variable. Why.
Ans. Velocity of an object changes when its magnitude or direction or both changes When the moon revolves round the earth, it is in circular motion and in a circular motion the direction changes at every point. Therefore the moon has variable velocity when it revolves round the earth.
12. When an athlete gives circular motion to a hammer or a discus and then releases it, the hammer or the discus moves in a straight line that is tangential to the circular path. Why.
Ans. Whenever a body is in circular motion, the direction of the motion changes at every point in its path. When the hammer or discus, moving in a circular path is released, it continues to move along the direction it has been moving at that particular point and instant. Therefore, the hammer or the discus moves in a direction tangential to the circular path
13. A particle is moving in a circle of diameter 5m. What is its displacement when it covers one and a half revolutions ?
Ans : After one and a half revolution , the particle reaches the diametrically opposite end . Thus , displacement = 5m .
14. Is it possible that the train in which you are sitting appears to move while it is at rest ?
Ans : The train in which we are sitting appears to move when the relative position of a point on adjacent train changes . This happens when we are at rest and adjacent train on next track starts moving .
15. The walls of your classroom are in motion but appear stationary . Explain .
Ans : The walls of my classroom are at rest with respect to us because their relative position remains constant . But to a person in outer space they appear moving as the earth rotates .
16. Why is the motion of a train starting from one station and stopping at the other is non-uniform ?
Ans : When the train starts from rest from a station , it accelerates to attain a maximum velocity . Thereafter , on reaching the next station , brakes are applied and it retards before it finally comes to rest . Thus , the motion of the train is non-uniform .
17. Define uniform acceleration .
Ans : The acceleration of a body is said to be uniform if its velocity changes by equal amount in equal intervals of time.
18. Define uniform motion .
Ans : When a body covers equal distances in equal intervals of time , then it has uniform motion .
19. Mohan travels at 20m/s from home to market and returns back at 25m/s . Find his average velocity for the entire journey .
Ans : Displacement of the Mohan over the journey = zero .
Thus , average velocity = zero .
20. What is uniform circular motion? Give one example.
Ans: When an object moves in a circular path with uniform speed its motion is called uniform circular motion. Eg: When an athlete throws a hammer or a discus in a sports meet, he / she holds the hammer or the discus in his / her hand and gives it a circular motion by rotating his / her own body. Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released.
21. Differentiate between velocity and acceleration.
Ans:
| Velocity | Acceleration |
| 1. The rate of change of position of body is called velocity. 2. S.I. unit is m/s. |
1. It is the rate of change of velocity. 2. S.I. unit is m/s2 . |
22. Differentiate between speed and velocity.
Ans: Speed is distance travelled per unit time and velocity is displacement per unit time.
23. Find the type of motion in the following case:
(a) A car travelling along a straight road, changes its speed.
(b) An athlete running a 100 m race.
(c) Moon revolving around the earth.
(d) An ant moving on the floor.
Ans:
(a) Non-uniform motion.
(b) Uniform motion.
(c) Uniform circular motion.
(d) Non-uniform motion.
24. It took 2 s after lightning for the sound of thunder to reach you. How far did the lightning struck?
Ans: Speed of sound is 346 m/s in air.
Time = 2 s
∴ Distance = ?
s = d / t
∴ d = s × t = 346 × 2 = 692 m.
∴ The lightning struck 692 m away.