Question:

A police van moving on a highway with a speed of 30 km/hr fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/hr. If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief’s car ? (Note:Obtain that speed which is relevant for damaging the thief’s car).

Answer:

Answer: Step 1: Calculate the relative velocity of the bullet with respect to the thief’s car. Relative velocity = (speed of police van + speed of bullet) - speed of thief’s car = (30 km/hr + 150 m/s) - 192 km/hr = (305/18 + 150) - 1925/18 = (250/18 + 150) - 960/18 = 400/18 m/s

Step 2: Convert the relative velocity into km/hr. 400/18 m/s = 400*18/1000 km/hr = 72 km/hr

Therefore, the bullet will hit the thief’s car with a speed of 72 km/hr, which is relevant for damaging the thief’s car.

Question:

On a two-lane road, car A is travelling with a speed of 36kmh^−1. Two cars B and C approach car A in opposite directions with a speed of 54kmh^−1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km,B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Answer:

1. Calculate the time taken for car B to travel 1 km from the point when the distance AB is equal to AC: Time taken = (1 km)/(54 km/h) = 0.0185 hours

2. Calculate the distance travelled by car A in the same time: Distance travelled by car A = (36 km/h) × (0.0185 hours) = 0.66 km

3. Calculate the minimum acceleration of car B required to avoid an accident: Minimum acceleration of car B = (1 km - 0.66 km)/(0.0185 hours) = 16.216 km/h^2

Question:

Explain clearly, with examples, the distinction between: (a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval. (b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].

Answer:

A) Magnitude of displacement (sometimes called distance) over an interval of time refers to the straight line distance between the initial and final positions of a particle over a given time interval. The total length of path covered by a particle over the same interval is the actual path that the particle has taken over the same time interval, which is usually longer than the magnitude of displacement. This is because the particle may have changed direction multiple times, resulting in a longer path than the direct line distance between the initial and final positions.

For example, if a particle moves from position A to position B in a straight line, the magnitude of displacement is the distance between A and B. However, if the particle moves from A to B by taking a zigzag path, the total length of path covered by the particle is greater than the magnitude of displacement.

The second quantity (total length of path covered by a particle) is always greater than the first (magnitude of displacement) as the total path length is always longer than the straight line distance between the initial and final positions.

B) Magnitude of average velocity over an interval of time refers to the average rate at which a particle is changing its position over a given time interval. The average speed over the same interval is the total path length divided by the time interval.

For example, if a particle moves from position A to position B in a straight line, the magnitude of average velocity is the rate at which the particle is changing its position over the given time interval. However, if the particle moves from A to B by taking a zigzag path, the average speed is greater than the magnitude of average velocity as the total path length is longer than the straight line distance between the initial and final positions.

The second quantity (average speed) is always greater than or equal to the first (magnitude of average velocity) as the total path length is always longer than or equal to the straight line distance between the initial and final positions.

The equality sign is true when the particle moves from A to B in a straight line, as the total path length is equal to the straight line distance between the initial and final positions.

Question:

Read each statement below carefully and state with reasons and examples, if it is true or false: A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant. (b) with zero speed may have non-zero velocity. (c) with constant speed must have zero acceleration. (d) with positive value of acceleration must be speeding up.

Answer:

(a) True. A particle in one-dimensional motion with zero speed at an instant may have non-zero acceleration at that instant. For example, if a particle is travelling in a straight line and is slowing down, it will have zero speed at the instant it stops but will have a negative acceleration at that instant.

(b) False. A particle with zero speed cannot have a non-zero velocity. Velocity is a vector quantity and is defined as the rate of change of displacement with respect to time. Since the particle has zero speed, its displacement is not changing, so its velocity is also zero.

(c) True. A particle with constant speed must have zero acceleration. This is because acceleration is defined as the rate of change of velocity with respect to time. Since the velocity is constant, it is not changing, so the acceleration must be zero.

(d) True. A particle with a positive value of acceleration must be speeding up. This is because acceleration is defined as the rate of change of velocity with respect to time. If the acceleration is positive, it means that the velocity is increasing, so the particle is speeding up.

Question:

A jet airplane travelling at the speed of 500kmh^−1 ejects its products of combustion at the speed of 1500kmh^−1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

Answer:

1. The speed of the products of combustion relative to the observer on the ground is 500kmh^−1 + 1500kmh^−1 = 2000kmh^−1.

2. Therefore, the speed of the jet plane with respect to an observer on the ground is 2000kmh^−1.

Question:

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward,followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Answer:

Step 1: The x-t graph of the motion of the drunkard can be plotted as follows:

x-axis: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13

t-axis: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21

Step 2: The drunkard takes 21 seconds to fall in the pit 13 m away from the start. This can be determined graphically by observing the x-t graph.

Step 3: The drunkard takes 21 seconds to fall in the pit 13 m away from the start. This can also be determined mathematically by calculating the total time taken by the drunkard to cover 13 m. The drunkard takes 5 steps forward and 3 steps backward in each cycle. Therefore, the total time taken by the drunkard to cover 13 m is 21 seconds (5+3+5+3+5+3+5+3+5+3+5+3+1 = 21).

Question:

A woman starts from her home at 9.00 am, walks with a speed of 5kmh^−1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25kmh^−1. Plot the position-time graph of the woman taking her home as origin.

Answer:

1. Draw the x-axis and the y-axis on the graph. Label the x-axis as ‘Time’ and the y-axis as ‘Position’.

2. Mark the origin on the graph. Label the origin as ‘Home’.

3. Mark the point where the woman reaches her office on the y-axis. Label this point as ‘Office’.

4. Calculate the time taken by the woman to reach her office from her home. It will take her 30 minutes (2.5 km/5 km/hr = 0.5 hr = 30 min).

5. Plot the point on the graph where the woman reaches her office. This point will be 30 minutes away from the origin on the x-axis and 2.5 km away from the origin on the y-axis.

6. Calculate the time taken by the woman to return home from her office. It will take her 12 minutes (2.5 km/25 km/hr = 0.1 hr = 12 min).

7. Plot the point on the graph where the woman returns home. This point will be 42 minutes away from the origin on the x-axis and 0 km away from the origin on the y-axis.

8. Connect the points plotted on the graph with a straight line. This will give the position-time graph of the woman taking her home as origin.

Question:

On a two lane road, car A is travelling with a speed of 36 km h^−1. two cars B and C approach car A in opposite directions with a speed of 54 km h^−1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident? A 9.8 m s^−2 B 10 m s^−2 C 1 m s^−2 D 2.0 m s−2

Answer:

Answer: A - 9.8 m s^−2

Question:

Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h−1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do buses ply on the road? A 9 min, 40 Kmph B 12 min, 10 Kmph C 12 min, 40 Kmph D 9 min, 60 Kmph

Answer:

Answer: C 12 min, 40 Kmph

Step 1: Consider the information given in the question. A man cycling with a speed of 20 km h−1 notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction.

Step 2: Calculate the speed of the bus in each direction. The speed of the bus in the direction of the man’s motion is 20 km/h ÷ 18 min = 1.11 km/min or 66.6 km/h. The speed of the bus in the opposite direction is 20 km/h ÷ 6 min = 3.33 km/min or 200 km/h.

Step 3: Since the speed of the bus is the same in both directions, the speed of the bus is 40 km/h.

Step 4: Calculate the period T of the bus service. The period T is the time it takes for a bus to travel from one town to the other. Since the speed of the bus is 40 km/h, the period T is 40 km/h ÷ 12 min = 3.33 km/min or 12 min.

Therefore, the period T of the bus service is 12 min and the speed of the buses (assumed constant) is 40 km/h. The correct answer is C 12 min, 40 Kmph.

Question:

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Answer:

1. At t = 0, the ball is dropped from a height of 90 m and has an initial speed of 0 m/s.

2. At t = 1 s, the ball has fallen a distance of 45 m and has a speed of 9 m/s.

3. At t = 2 s, the ball has fallen a distance of 81 m and has a speed of 8.1 m/s.

4. At t = 3 s, the ball has fallen a distance of 108.9 m and has a speed of 7.29 m/s.

5. At t = 4 s, the ball has fallen a distance of 126.21 m and has a speed of 6.561 m/s.

6. At t = 5 s, the ball has fallen a distance of 135.55 m and has a speed of 5.899 m/s.

7. At t = 6 s, the ball has fallen a distance of 144.2 m and has a speed of 5.319 m/s.

8. At t = 7 s, the ball has fallen a distance of 152.18 m and has a speed of 4.797 m/s.

9. At t = 8 s, the ball has fallen a distance of 159.5 m and has a speed of 4.325 m/s.

10. At t = 9 s, the ball has fallen a distance of 166.15 m and has a speed of 3.892 m/s.

11. At t = 10 s, the ball has fallen a distance of 172.14 m and has a speed of 3.493 m/s.

12. At t = 11 s, the ball has fallen a distance of 177.43 m and has a speed of 3.128 m/s.

13. At t = 12 s, the ball has fallen a distance of 182.1 m and has a speed of 2.789 m/s.

The speed-time graph of the ball’s motion is as follows:

Speed (m/s)

Question:

A jet aeroplane travelling at the speed of 500km/hr ejects its products of combustion at speed of 1500km/hr relative to jet plane. What is the relative velocity of the latter with respect to an observer on the ground?

Answer:

1. The relative velocity of the jet plane with respect to an observer on the ground is 500km/hr.

2. The relative velocity of the products of combustion with respect to the observer on the ground is 500km/hr + 1500km/hr = 2000km/hr.

Question:

A car moving along a straight highway with speed of 126Km/h is brought to a stop within a distance of 200m. What is the retardation of the car (assumed uniform ) and how long does it take for the car to stop? A 3.06m/s2 and 11.4s B 2.06m/s2 and 11.4s C 3.06m/s2 and 10.4s D 3.06m/s2 and 4.1s

Answer:

Answer: D 3.06m/s2 and 4.1s

Step 1: Calculate the acceleration of the car.

Acceleration = (Final Velocity - Initial Velocity) / Time

Time = Distance / Velocity

Time = 200m / 126Km/h

Time = 0.001667 hours

Acceleration = (0 - 126Km/h) / 0.001667 hours

Acceleration = -75600Km/h2

Acceleration = -75600m/s2

Acceleration = -75.6m/s2

Step 2: Calculate the retardation of the car.

Retardation = -75.6m/s2

Step 3: Calculate the time taken for the car to stop.

Time = Distance / Velocity

Time = 200m / (126Km/h + 75.6m/s2)

Time = 200m / (126Km/h + 75.6m/s2)

Time = 4.1s

Question:

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^−1 . Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^−1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min (iii) 0 to 40 min?

Answer:

(a) i) 0 to 30 min: Average velocity = 0 km/h

ii) 0 to 50 min: Average velocity = (2.5 km/h + (-2.5 km/h))/2 = 0 km/h

iii) 0 to 40 min: Average velocity = (2.5 km/h + (-2.5 km/h))/2 = 0 km/h

(b) i) 0 to 30 min: Average speed = 5 km/h

ii) 0 to 50 min: Average speed = (2.5 km/h + 7.5 km/h)/2 = 5 km/h

iii) 0 to 40 min: Average speed = (2.5 km/h + 7.5 km/h)/2 = 5 km/h

Question:

In which of the following examples of motion, can the body be considered approximately a point object: (a) A railway carriage moving without jerks between two stations. (b) A monkey sitting on top of a man cycling smoothly on a circular track. (c) A spinning cricket ball that turns sharply on hitting the ground. (d) A tumbling beaker that has slipped off the edge of a table.

Answer:

Answer: (a) A railway carriage moving without jerks between two stations. (b) A monkey sitting on top of a man cycling smoothly on a circular track.

Question:

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed =49m/s. a) How much time does the ball take to return to his hands? b) If the lift starts moving up with a uniform speed of 5m/s and he again throws the ball , how long does it take to return to his hands ?

Answer:

a) The time taken for the ball to return to the boy’s hands is given by:

t = 2v/g

where v is the initial velocity of the ball (49 m/s) and g is the acceleration due to gravity (9.8 m/s2).

Therefore, t = 2(49)/9.8 = 10 s.

b) The time taken for the ball to return to the boy’s hands when the lift is moving up with a uniform speed of 5 m/s is given by:

t = 2(v+u)/g

where v is the initial velocity of the ball (49 m/s), u is the uniform speed of the lift (5 m/s) and g is the acceleration due to gravity (9.8 m/s2).

Therefore, t = 2(49+5)/9.8 = 11 s.

Question:

A player throws a ball upwards with an initial speed of 29.4ms^−1. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the players hands ? (Take g=9.8m/s^2 and neglect air resistance).

Answer:

(a) The direction of acceleration during the upward motion of the ball is in the direction of gravity, i.e. downwards.

(b) At the highest point of its motion, the velocity of the ball is 0 ms^−1 and the acceleration of the ball is -9.8 ms^−2 (downwards).

(c) During the upward motion of the ball, the position is in the positive direction of the x-axis, the velocity is in the positive direction of the x-axis, and the acceleration is in the negative direction of the x-axis. During the downward motion of the ball, the position is in the negative direction of the x-axis, the velocity is in the negative direction of the x-axis, and the acceleration is in the positive direction of the x-axis.

(d) The ball rises to a height of 8.9 m and returns to the player’s hands after 0.9 s.

Question:

We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Answer:

1. Average speed is the total distance traveled over a period of time, while average velocity is the displacement of an object over a period of time.

2. Instantaneous speed is the speed of an object at a given moment in time, while instantaneous velocity is the displacement of an object at a given moment in time.

3. Since instantaneous speed is the speed of an object at a single point in time, it is always equal to the magnitude of instantaneous velocity, which is the displacement of an object at a single point in time.