Q 1 How do we often perceive an object to be in motion?
Ans: When its position changes with time.
Q 2. What is essential to describe the position of an object?
Ans: a reference point, called the origin.
Q 3. Give the technical term used for the rate of motion of a body. Give its SI unit.
Ans: Speed. metre per second (m/s)
Q 4. A truck is moving on a straight road with uniform acceleration. The following table gives the speed of the truck at various instants of time :
Speed
(m/s) 5 10 15 20 25 30
Time (s) 0 10 20 30 40 50
i. Draw a speed – time graph by choosing a convenient scale. Determine from it :
The acceleration of truck.
ii. The distance travelled by the truck in 50 seconds. Ans: Draw graph with x- axis and y – axis on the graph ,
1cm = 10 s on x – axis.
1cm = 5m on y – axis.
i. Acceleration means slope of graph AB,
= = 0.5 m/s
𝐀𝐂 𝟓𝟎− 𝟎
ii. Distance travelled by truck = area of trapezium OABD
𝟏
= ( AO + BD) × OD
𝟐
= 875 m
Q 5. An object travels a distance of 16 m in 4s and then another 16m in 2 s. What is the average speed of the object?
Ans: 𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞
Average speed =
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞 𝟑𝟐
= = 5.33 m/s
𝟔
Q 6. Name the physical quantities denoted by :
i. The slope of the distance – time graph
Ans: Speed
ii. The area under velocity – time graph.
Ans: Distance
Q 7. In which direction do the following forces act when object is in motion :
𝐁𝐂 𝟑𝟎−𝟓 2
i. Frictional force.----------Backward ii. Gravitational force.-----Downward iii. Centripetal force.------------Towards the centre.
Q 8. Define average speed and average velocity.
Ans: Average speed: It is obtained by total distance travelled by total time
taken.
Average velocity: It is arithmetic mean of initial and final velocity.
Q 9. A train travels at a speed of 60 km/hr for 0.52 hr, at 30 km/hr for the next 0.24 hr and then, at 70 km/hr for the next 0.71 hr. What is the average speed of the train?
Ans: In first case, the train travels at a speed of 60 km/hr for a time of 0.52 h
Distance = 31.2 km.................(i)
In second case distance = 7.2 km ..........(ii) in third case , distance = 49.7 km ..........(iii)
From (i) , (ii) and (iii),
Total distance travelled = 88.1 km total time taken = 1.47 h
Average speed = 59.9 km
Q 10. Define the following terms:
i. Acceleration
Ans: Acceleration is the rate of change of velocity of an object with respect to time.
ii. Uniform Motion
Ans: When body covers equal distance in equal interval of time.
iii. Uniform accelerated motion
Ans: A body is said to have uniform accelerated motion if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
Q 11. What do the following instrument measures? i. Odometer – distance ii. Speedometer – instantaneous speed
Q 12. Plot velocity-time graph of a body whose initial velocity is 5 m/s and is moving with retardation of 1m/s2.Also calculate the distance covered by it.
Ans:
𝟏
Distance = area under the graph = × 𝒃𝒂𝒔𝒆 × 𝒉𝒆𝒊𝒈𝒉𝒕
𝟐
𝟏
= × 𝟓 × 𝟓
𝟐
=12.5 m
Q 13. Study the velocity-time graph of a car given below and answer the following questions:
iv. What type of motion is represented by OA?
v. Find the acceleration from B to C.
vi. Calculate the distance covered by the body from A to B.
Ans: i. Uniform acceleration.
ii. 𝒂 = 𝒗−𝒖 = 𝟎−𝟑 = −𝟎. 𝟏𝟓 𝒎/𝒔𝟐
𝒕 𝟐𝟎
iii. The area under graph gives distance.
Distance from A to B = speed x time
= 3 x (60-20)
= 120 m
Q 14. Write three advantages of velocity-time graph.
Ans: • Instantaneous velocity of the moving body at any period of time can be determined.
• Displacement of the body can be determined.
• Acceleration of the body can be determined.
Q 15. Study the velocity-time graph of a body given below and answer the following questions:
i. State the type of motion
represented by part BC.
ii. Identify the part of graph which has zero acceleration. Give reason for your answer.
Ans: i. Uniform motion ii. Part BC because velocity remains constant from B to C.
Q 16. State with reasons, if it is possible or impossible for an object to have acceleration opposite to the direction of motion.
Ans: Possible
because in negative acceleration, acceleration is against the motion of the body.
Q 17. A bus accelerates from rest at the rate of 0.2 m/s2. What will be the velocity attained by it in 1 minute?
Ans: t = 1 min = 60 seconds u = 0 v = u + at v = 0 + (0.2 x 60) v = 12 m/s
Q 18. A Policeman ‘P’ is chasing a thief ‘T’. The distance-time graph of their motion is given below.
i. How much is T ahead of P when
the motion starts?
ii. When and where will P catch T?
Ans: i. 40 m
ii. P catches T at Q, 60 metres from origin, after 3 seconds.
Q 19. Define Uniform Circular Motion.
Ans: When a body moves in a circular path with uniform speed , its motion is called uniform circular motion.
Q 20. What does the area under velocity – time graph represent?
Ans: Displacement
Q 21. Differentiate between:
Uniform motion in a straight line and Uniform circular motion Ans:
Uniform motion in a straight line Uniform circular motion
1. The body moves along a straight path
2. The direction of motion does not change
3. It is not an accelerated motion 1. The body moves along a circular path
2. The direction of motion changes continuously
3. It is an accelerated motion
Q 22. Using the graph given below, derive II equation of motion for position-time relation.
Ans: Distance travelled (s) = area of trapezium OABC
∴ 𝐬 = 𝐚𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐁𝐃 + 𝐚𝐫𝐞𝐚 𝐨𝐟 □ 𝐎𝐀𝐃𝐂
𝟏
∴ 𝐬 = (× 𝐛𝐚𝐬𝐞 × 𝐡𝐞𝐢𝐠𝐡𝐭) + (𝐥𝐞𝐧𝐠𝐭𝐡 × 𝐛𝐫𝐞𝐚𝐝𝐭𝐡)
𝟐
𝟏
∴ 𝐬 = (× 𝐀𝐃 × 𝐁𝐃) + (𝐎𝐂 × 𝐎𝐀)
𝟐
𝟏
∴ 𝐬 = (× 𝐭 × 𝐁𝐃) + (𝐭 × 𝐮) ....(1)
𝟐
But BD = BC – DC
= v – u
= (u + at) –u
= at
Substituting BD = at in (1), we get
𝟏
𝐬 = ( × 𝐭 × 𝐚𝐭) + (𝐭 × 𝐮)
𝟐
𝟏
𝐬 = 𝐮𝐭 + 𝐚𝐭𝟐
𝟐
Q 23. What is negative acceleration? Give one example of it.
Ans: • If the velocity of a body decreases with time, it is said to have negative acceleration.
• The direction of negative acceleration is opposite to the direction of motion/velocity.
• Example: When the driver of the moving car applies brakes and the car stops, the car possesses negative acceleration. (example may vary, accept all correct examples)
Q 24. A ball is thrown vertically upward with a velocity of 5 m/s. It uniformly accelerates at the rate of -10m/s2, stops after reaching a certain height and begins to fall back downwards. Calculate the height attained by the ball before it falls back.
Ans: 𝟐𝐚𝐬 = 𝐯𝟐 − 𝐮𝟐
∴ 𝟐 × (−𝟏𝟎) × 𝐬 = 𝟎𝟐 − 𝟓𝟐
−𝟐𝟓
∴ 𝐬
∴ 𝐬 = 𝟏. 𝟐𝟓 𝐦
Q 25. Study the graph given below and answer the questions based on it:
i. State the type of motion represented by the part AB.
ii. Calculate the distance travelled by the object in the first five hours.
Ans: i. Uniform accelerated motion
(Note: ‘retardation’ should not be considered as answer because the question asks the ‘type of motion’ and not the ‘type of acceleration’.)
ii. 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐭𝐫𝐚𝐯𝐞𝐥𝐥𝐞𝐝 𝐢𝐧 𝐟𝐢𝐫𝐬𝐭 𝐟𝐢𝐯𝐞 𝐡𝐨𝐮𝐫𝐬 = 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐎𝐂𝐀
𝟏
= × 𝐛𝐚𝐬𝐞 × 𝐡𝐞𝐢𝐠𝐡𝐭
𝟐
𝟏
= × 𝟓 × 𝟏𝟎
𝟐
= 𝟐𝟓 𝐤𝐦
Q 26. A car covered 85 km in the first two hours and 150 km in the next three hours.
Calculate the average speed of the car.
Ans 𝐭𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞
𝐚𝐯𝐞𝐫𝐚𝐠𝐞 𝐬𝐩𝐞𝐞𝐝 =
𝐭𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
𝟖𝟓 + 𝟏𝟓𝟎
∴ 𝐚𝐯𝐞𝐫𝐚𝐠𝐞 𝐬𝐩𝐞𝐞𝐝 =
𝟐 + 𝟑
∴ 𝐚𝐯𝐞𝐫𝐚𝐠𝐞 𝐬𝐩𝐞𝐞𝐝 = 𝟒𝟕 𝐤𝐦/𝐡𝐫
Q 27. A mechanic strikes a nail with hammer moving with a velocity of 20 m/s. The hammer comes to rest in 0.02 s after striking the nail. Calculate the acceleration of the nail.
𝐯 − 𝐮
Ans 𝐚 =
𝐭
𝟎 − 𝟐𝟎
∴ 𝐚
∴ 𝐚 = −𝟏𝟎𝟎𝟎 𝐦/𝐬𝟐
Q 28. The driver of a train which is moving with a velocity of 108 km/h applies brakes. The train retards uniformly at the rate of -1 m/s2 and stops after covering certain distance. Find the distance covered by the train during retardation. Ans When the train comes to rest,
𝐯 = 𝟎 m/s, 𝐚 = −𝟏𝐦/𝐬𝟐, 𝐮 = 𝟏𝟎𝟖𝐤𝐦/𝐡𝐫 = 𝟏𝟎𝟖 ×𝟏𝟎𝟎𝟎 = 𝟑𝟎 𝐦/𝐬
𝟑𝟔𝟎𝟎
𝟐𝐚𝐬 = 𝐯𝟐 − 𝐮𝟐
∴ 𝟐 × (−𝟏) × 𝐬 = 𝟎𝟐 − (𝟑𝟎)𝟐
∴ 𝐬 = 𝟒𝟓𝟎 𝐦
0 Comments