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Q1. What is the physical quantity that gives the location of an object relative to a reference point?
Position is a physical quantity that tells us exactly where an object is located with respect to a fixed reference point, such as the origin on a coordinate system. Distance and displacement describe how far an object has moved, while speed describes how fast it is moving.
Q2. A man walks 10 m east and then 7 m west. What is his final position relative to his starting point?
The man first moves 10 m in the east direction and then moves 7 m in the opposite direction (west). To find the final position, we subtract the westward distance from the eastward distance: 10 m – 7 m = 3 m. Since the larger distance was towards the east, his final position is 3 m east of the starting point.
Q3. If an object’s displacement is zero, what can you surely say about the distance it traveled?
Displacement is zero when the object returns to its starting point, regardless of the path taken. However, the distance traveled is the total length of the actual path, which must be greater than zero if the object actually moved. For example, if you walk around a park and come back to the same spot, displacement is zero but distance is the full length of the path you walked.
Q4. What is the speed of an object that covers 2.5 kilometres in 30 minutes?
Speed is calculated as distance divided by time. Here distance is 2.5 km. Time is 30 minutes, which is 0.5 hours. So speed = 2.5 km ÷ 0.5 h = 5 km/h. This means the object covers 5 kilometres every hour.
Q5. An object moves with a constant speed of 5 m/s. What does this tell you about its velocity?
Speed is the magnitude of velocity. If only speed is constant, it does not guarantee that velocity is constant because velocity also depends on direction. The object could be changing direction while maintaining the same speed, like in circular motion. So we can only say the magnitude of velocity is constant.
Q6. A bird flies north at 15 m/s for 4 seconds, then instantly turns and flies south at 15 m/s for 4 seconds. What is its average speed?
Average speed is total distance divided by total time. Distance covered in the first part = 15 × 4 = 60 m. Distance in the second part = 15 × 4 = 60 m. Total distance = 60 + 60 = 120 m. Total time = 4 + 4 = 8 s. Average speed = 120 ÷ 8 = 15 m/s.
Q7. For the bird in the previous question, what is the magnitude of its average velocity?
Average velocity is total displacement divided by total time. The bird flies 60 m north and then 60 m south, so it returns to its starting point. The displacement is zero because the initial and final positions are the same. Therefore, average velocity = 0 ÷ 8 = 0 m/s.
Q8. What can you say about the acceleration of an object in uniform motion?
Uniform motion means the object moves with a constant velocity, meaning both speed and direction are constant. Since velocity is not changing, the rate of change of velocity, which is acceleration, must be zero.
Q9. A car’s odometer reads 2500 km at the start of a trip and 2850 km at the end. If the trip took 5 hours, what was the average speed?
The distance traveled is the difference in odometer readings: 2850 km – 2500 km = 350 km. The time taken is 5 hours. Average speed = distance ÷ time = 350 ÷ 5 = 70 km/h.
Q10. If the distance-time graph of an object is a curve that gets steeper, the object is ______.
The slope of a distance-time graph represents speed. If the curve gets steeper, it means the slope is increasing, which indicates that the speed of the object is increasing over time. Therefore, the object is speeding up.
Q11. A scooter’s velocity changes from 18 km/h to 36 km/h in 10 seconds. What is its acceleration in m/s²?
First convert velocities from km/h to m/s. 18 km/h = 18 × (5/18) = 5 m/s. 36 km/h = 36 × (5/18) = 10 m/s. Acceleration = (final velocity – initial velocity) ÷ time = (10 – 5) ÷ 10 = 5 ÷ 10 = 0.5 m/s².
Q12. A ball is thrown vertically upwards. What is its acceleration at the very highest point of its journey? (Ignore air resistance)
Throughout the entire journey of a freely falling object (including when thrown upwards), the acceleration due to gravity is always constant and acts downwards towards the Earth. Even at the highest point where the velocity becomes zero for an instant, the acceleration remains 9.8 m/s² downwards.
Q13. Which equation would you use to find the time taken, if you know initial velocity, final velocity, and acceleration?
The first equation of motion, v = u + at, relates initial velocity (u), final velocity (v), acceleration (a), and time (t). This equation can be rearranged to find time: t = (v – u)/a. The other equations do not directly give time from just these three quantities.
Q14. A bus decelerates uniformly from 72 km/h to a stop in 10 seconds. How far does it travel while stopping?
Convert 72 km/h to m/s: 72 × (5/18) = 20 m/s. Initial velocity u = 20 m/s, final velocity v = 0 m/s, time t = 10 s. Using s = (u+v)/2 × t = (20+0)/2 × 10 = 10 × 10 = 100 m.
Q15. On a velocity-time graph, a straight line sloping downwards to the right indicates ______.
A downward sloping line on a velocity-time graph means velocity is decreasing uniformly with time. This indicates a constant negative acceleration, which is also called retardation or deceleration. The object is slowing down at a uniform rate.
Q16. What does a point on a velocity-time graph directly represent?
On a velocity-time graph, the y-axis represents velocity. Each point on the graph shows the velocity of the object at a specific instant of time. The slope gives acceleration, and the area under the curve gives displacement.
Q17. The area under an acceleration-time graph represents ______.
The area under an acceleration-time graph gives the change in velocity of the object. This is because acceleration is the rate of change of velocity. Multiplying acceleration by time gives the change in velocity.
Q18. In uniform circular motion, what provides the necessary centripetal force for a stone tied to a string?
When a stone is tied to a string and whirled in a circle, the string pulls the stone towards the centre. This pulling force is called tension. This tension provides the centripetal force required to keep the stone moving in a circular path.
Q19. A passenger sitting in a moving train observes that the trees outside are moving backwards. This is an example of ______.
The trees are actually stationary, but they appear to move backwards because the passenger is in a moving train. This shows that motion depends on the observer’s frame of reference. This is called relative motion, where the motion of an object is described with respect to another object.
Q20. If two cars, A and B, are moving in the same direction at 50 km/h and 70 km/h respectively, what is the velocity of car B as seen from car A?
Relative velocity of B with respect to A is given by velocity of B minus velocity of A. So, 70 km/h – 50 km/h = 20 km/h. Since the result is positive and they are moving in the same direction, car B appears to move forward at 20 km/h when observed from car A.
Q21. An object moves 200 metres in 40 seconds. Calculate its speed.
Speed is distance divided by time. Distance = 200 m, time = 40 s. Speed = 200 ÷ 40 = 5 m/s. This means the object covers 5 metres every second.
Q22. A cyclist covers a distance of 1.5 km in 5 minutes. Find his speed in m/s.
Convert distance to metres: 1.5 km = 1500 m. Convert time to seconds: 5 minutes = 5 × 60 = 300 seconds. Speed = 1500 ÷ 300 = 5 m/s.
Q23. A train 150 m long is moving with a constant velocity of 72 km/h. How long will it take to cross a 250 m long platform completely?
To completely cross the platform, the train must cover its own length plus the length of the platform. Total distance = 150 + 250 = 400 m. Convert speed to m/s: 72 km/h = 72 × (5/18) = 20 m/s. Time = distance ÷ speed = 400 ÷ 20 = 20 seconds.
Q24. A body starting from rest accelerates uniformly at 4 m/s². Calculate its velocity after 6 seconds.
Using the first equation of motion, v = u + at. Here initial velocity u = 0 (starts from rest), acceleration a = 4 m/s², time t = 6 s. So v = 0 + 4 × 6 = 24 m/s.
Q25. A car moving at 15 m/s applies brakes and comes to rest in 3 seconds. Calculate its retardation (deceleration).
Acceleration = (final velocity – initial velocity) ÷ time. Initial velocity u = 15 m/s, final velocity v = 0 m/s, time t = 3 s. a = (0 – 15) ÷ 3 = -15 ÷ 3 = -5 m/s². The negative sign indicates retardation, so the magnitude of retardation is 5 m/s².
Q26. Using the data from the previous question (car decelerating at 5 m/s²), calculate the distance it travels before stopping.
Using the third equation of motion, v² = u² + 2as. Here u = 15 m/s, v = 0 m/s, a = -5 m/s². Substituting: 0 = (15)² + 2(-5)s => 0 = 225 – 10s => 10s = 225 => s = 22.5 m.
Q27. A body is thrown vertically upwards with a velocity of 49 m/s. Taking g = 9.8 m/s², calculate the time it takes to reach the maximum height.
At maximum height, final velocity v = 0. Using v = u + at, where a = -g = -9.8 m/s² (negative because acceleration is downwards). 0 = 49 – 9.8t => 9.8t = 49 => t = 49 ÷ 9.8 = 5 seconds.
Q28. For the body thrown upwards in Q27, what is the maximum height reached?
Using v² = u² + 2as, where v = 0, u = 49 m/s, a = -9.8 m/s². 0 = 49² + 2(-9.8)s => 0 = 2401 – 19.6s => 19.6s = 2401 => s = 2401 ÷ 19.6 = 122.5 m.
Q29. A particle moving with an initial velocity of 2 m/s is subjected to a constant acceleration of 1.5 m/s² for 8 seconds. Find the distance traveled.
Using the second equation of motion, s = ut + ½at². Here u = 2 m/s, a = 1.5 m/s², t = 8 s. s = 2(8) + ½(1.5)(8)² = 16 + 0.5 × 1.5 × 64 = 16 + 48 = 64 m.
Q30. A bus increases its speed from 36 km/h to 54 km/h in 5 seconds. Calculate the distance covered during this acceleration.
Convert speeds to m/s: 36 km/h = 10 m/s, 54 km/h = 15 m/s. Using v² = u² + 2as: (15)² = (10)² + 2a(5). First find a = (15-10)/5 = 1 m/s². Then s = ut + ½at² = 10(5) + 0.5(1)(25) = 50 + 12.5 = 62.5 m.
Q31. An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed for the entire journey?
Total distance = 16 + 16 = 32 m. Total time = 4 + 2 = 6 s. Average speed = total distance ÷ total time = 32 ÷ 6 = 5.33 m/s.
Q32. A body moves with a velocity of 10 m/s for 5 seconds. It then accelerates uniformly for the next 10 seconds to reach a velocity of 30 m/s. Find the total distance covered.
Distance in first part = velocity × time = 10 × 5 = 50 m. For second part: u = 10 m/s, v = 30 m/s, t = 10 s. Using s = (u+v)/2 × t = (10+30)/2 × 10 = 20 × 10 = 200 m. Total distance = 50 + 200 = 250 m.
Q33. A train accelerates from rest at a constant rate α for some time, then decelerates at a constant rate β to come to rest. If the total time of journey is T, what is the maximum velocity attained?
Let t1 be time of acceleration and t2 be time of deceleration. Then t1 + t2 = T. Maximum velocity v = αt1 = βt2. So t1 = v/α and t2 = v/β. Substituting: v/α + v/β = T => v(1/α + 1/β) = T => v(α+β)/(αβ) = T => v = (αβT)/(α+β).
Q34. The speed of a car as a function of time is given by v = 5t + 3, where v is in m/s and t in seconds. What is the acceleration of the car?
Acceleration is the rate of change of velocity with respect to time. In the equation v = 5t + 3, the coefficient of t is 5, which represents the acceleration. So the acceleration is constant at 5 m/s².
Q35. Using the velocity function from Q34 (v=5t+3), find the distance traveled between t=2s and t=4s.
Distance is the integral of velocity with respect to time. Distance = ∫(5t+3)dt from 2 to 4 = [5t²/2 + 3t] from 2 to 4. At t=4: 5(16)/2 + 12 = 40 + 12 = 52. At t=2: 5(4)/2 + 6 = 10 + 6 = 16. Distance = 52 – 16 = 36 m. (Note: Correct calculation gives 36 m, not 42 m).
Q36. The displacement (s) of a particle is given by s = 2t² + 5t + 10, where s is in metres and t in seconds. What is its initial velocity?
Velocity is the derivative of displacement with respect to time. v = ds/dt = 4t + 5. Initial velocity means velocity at t = 0. So v(0) = 4(0) + 5 = 5 m/s.
Q37. From the displacement function in Q36 (s=2t²+5t+10), what is the acceleration of the particle?
Acceleration is the derivative of velocity with respect to time. From v = 4t + 5, a = dv/dt = 4 m/s². This is the coefficient of t² in the displacement equation multiplied by 2.
Q38. A ball is dropped from a height of 80 m. How long does it take to hit the ground? (Take g = 10 m/s²)
Using s = ut + ½at², where u = 0 (dropped from rest), s = 80 m, a = g = 10 m/s². 80 = 0 + ½(10)t² => 80 = 5t² => t² = 16 => t = 4 seconds.
Q39. A stone is projected upwards with a speed of 40 m/s. How long will it take to return to the point of projection? (g=10 m/s²)
Time to reach maximum height is t = u/g = 40/10 = 4 s. The time to go up and come back down is twice this value because the motion is symmetric. Total time = 2 × 4 = 8 seconds.
Q40. Two trains of lengths 100 m and 150 m are moving towards each other at speeds of 54 km/h and 72 km/h respectively. Find the time taken to cross each other completely.
Total distance to be covered = 100 + 150 = 250 m. Convert speeds to m/s: 54 km/h = 15 m/s, 72 km/h = 20 m/s. Relative speed when moving towards each other = 15 + 20 = 35 m/s. Time = distance ÷ relative speed = 250 ÷ 35 = 7.14 s (approximately 7.5 s).
Q41. Can the average speed of a moving body ever be zero?
Average speed is total distance divided by total time. For a moving body, the total distance covered is always greater than zero (unless it never moves). Therefore, average speed can never be zero for a moving body. Average velocity, however, can be zero if displacement is zero.
Q42. Which of the following statements is true for a particle moving in a straight line with constant speed?
In a straight line, if speed is constant, the direction does not change. Since both speed and direction are constant, the velocity is constant. Therefore, acceleration is zero because there is no change in velocity.
Q43. A particle covers half of its total distance with speed v1 and the other half with speed v2. What is its average speed for the entire journey?
Let the total distance be 2s. Time for first half = s/v1, time for second half = s/v2. Total time = s(1/v1 + 1/v2) = s(v1+v2)/(v1v2). Average speed = total distance ÷ total time = 2s ÷ [s(v1+v2)/(v1v2)] = (2v1v2)/(v1+v2).
Q44. For motion along a straight line, if the velocity and acceleration have the same sign, the object is ______.
If velocity and acceleration have the same sign, it means the acceleration is acting in the same direction as the motion. This causes the speed of the object to increase. If they have opposite signs, the object is slowing down.
Q45. If the velocity-time graph is a curve concave upwards, the acceleration is ______.
The slope of a velocity-time graph gives acceleration. A curve that is concave upwards means the slope is increasing over time, so the acceleration is increasing. If the slope is constant, the graph would be a straight line.
Q46. The motion of the moon around the earth is approximately ______.
The moon moves around the earth in a nearly circular orbit with approximately constant speed. Since the path is circular and the speed is nearly constant, this is an example of uniform circular motion.
Q47. What does the odometer of a car measure?
An odometer is a device in a vehicle that records the total distance traveled by the vehicle since it was manufactured or since the last reset. It does not measure speed or displacement.
Q48. A particle moves along a semicircle of radius R in time T. What is the magnitude of its average velocity?
Average velocity is displacement divided by time. For a semicircle, the displacement is the straight-line distance between the starting and ending points, which is the diameter of the circle (2R). So average velocity = 2R/T.
Q49. If the position-time graph of an object is a straight line parallel to the position axis, it indicates the object is ______.
A straight line parallel to the position axis (vertical line) on a position-time graph means that the time is changing but the position is not. This indicates the object is not moving, meaning it is at rest.
Q50. The rate of change of displacement with respect to time is ______.
Velocity is defined as the rate of change of displacement with respect to time. It tells us how quickly the position of an object is changing. Speed is the rate of change of distance, while acceleration is the rate of change of velocity.