Chapter 2: Motion in a Straight Line Application Questions | physics Class 11 CBSE

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Chapter 2: Motion in a Straight Line Application Questions | physics Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A car travelling at 126 km h⁻¹ is brought to rest within a distance of 200 m by uniform braking. Calculate the retardation of the car and the time taken to stop.

Convert initial speed: v0 = 126 km h⁻¹ = 126 × (1000/3600) = 35 m s⁻¹. Final velocity v = 0 m s⁻¹.
Using v² = v0² + 2ax: 0 = (35)² + 2a(200), so 2a(200) = -1225, giving a = -3.0625 ≈ -3.06 m s⁻² (retardation = 3.06 m s⁻²).
Using v = v0 + at: 0 = 35 + (-3.0625)t, giving t = 35/3.0625 ≈ 11.4 s.
This shows that higher speeds require greater retardation or longer distances to stop safely — a direct application of kinematic equations to road safety.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A player throws a ball vertically upward at 29.4 m s⁻¹. Using g = 9.8 m s⁻², find (a) the maximum height reached and (b) the time after which it returns to the player's hands.

At maximum height, v = 0. Using v² = v0² - 2gh: 0 = (29.4)² - 2(9.8)h, so h = (29.4)²/19.6 = 864.36/19.6 = 44.1 m.
Time to reach maximum height: using v = v0 - gt: 0 = 29.4 - 9.8t, so t1 = 3 s. By symmetry, total time of flight = 2t1 = 6 s.
Alternatively, using y = v0t - ½gt² = 0 for return: t(v0 - ½gt) = 0, giving t = 2v0/g = 2 × 29.4/9.8 = 6 s.
These results illustrate the symmetry of vertical projectile motion under constant gravitational acceleration — time of ascent equals time of descent.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A police van moving at 30 km h⁻¹ fires a bullet with muzzle speed 150 m s⁻¹ at a thief's car moving at 192 km h⁻¹ in the same direction. Find the speed of the bullet relative to the thief's car.

Convert: speed of police van = 30 km h⁻¹ = 30/3.6 ≈ 8.33 m s⁻¹; speed of thief's car = 192/3.6 ≈ 53.33 m s⁻¹.
Speed of bullet relative to ground = speed of van + muzzle speed = 8.33 + 150 = 158.33 m s⁻¹.
Speed of bullet relative to thief's car = 158.33 - 53.33 = 105 m s⁻¹.
This is the relative velocity that determines the damaging impact on the thief's car; the concept of relative velocity is essential for such problems involving multiple moving objects.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A woman walks from home to her office 2.5 km away at 5 km h⁻¹, stays all day, then returns by auto at 25 km h⁻¹. For the entire journey, find (a) average velocity and (b) average speed.

Since she returns to her starting point (home), the net displacement = 0 km. Therefore, average velocity for the entire trip = 0 km h⁻¹.
Time to walk to office = 2.5/5 = 0.5 h; time to return by auto = 2.5/25 = 0.1 h. Total travel time = 0.6 h.
Total distance = 2.5 + 2.5 = 5 km. Average speed = 5/0.6 ≈ 8.33 km h⁻¹.
This example demonstrates why average speed and average velocity must be defined differently — average velocity is zero for a round trip, which would misleadingly suggest the person never moved.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

In a reaction time experiment, a ruler is dropped and falls 21 cm before being caught. Estimate the reaction time using g = 9.8 m s⁻².

The ruler undergoes free fall from rest (v0 = 0), so d = ½g·tr², where d is the distance fallen and tr is the reaction time.
Rearranging: tr = √(2d/g) = √(2 × 0.21/9.8) = √(0.042/9.8) = √(0.04286) ≈ 0.207 s.
Therefore, reaction time ≈ 0.2 s, which represents the time the person takes to observe, process, and respond to the ruler being dropped.
This experiment illustrates practical kinematics: free fall equations directly quantify a human physiological parameter, demonstrating the broad applicability of kinematic principles.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

With reference to a freely falling body starting from rest, show that the distances covered in the 1st, 2nd, and 3rd seconds are in the ratio 1:3:5.

Using y = ½gt² (taking downward as positive, starting from rest), distance after 1 s: y1 = ½g(1)² = ½g; after 2 s: y2 = ½g(4) = 2g; after 3 s: y3 = ½g(9) = 4.5g.
Distance in 1st second = ½g; in 2nd second = 2g - ½g = (3/2)g; in 3rd second = (9/2)g - 2g = (5/2)g.
Ratio of distances: ½g : (3/2)g : (5/2)g = 1:3:5, confirming Galileo's law of odd numbers for the first three intervals.
This is a direct consequence of uniform acceleration — because velocity increases linearly with time, the distances covered in successive equal intervals increase by equal odd-number increments.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A drunkard in a narrow lane takes 5 steps forward and 3 steps backward repeatedly, with each step being 1 m and taking 1 s. Determine after how many seconds he falls into a pit 13 m away.

In one complete cycle (5 steps forward + 3 steps backward = 8 steps, 8 s), the net displacement = 5 - 3 = 2 m.
After 4 complete cycles (32 s), net displacement = 4 × 2 = 8 m. At the start of the 5th cycle, he takes 5 forward steps, reaching 8 + 5 = 13 m at step 37 (i.e., 37 s).
Therefore, the drunkard falls into the pit after 37 s from the start.
This problem illustrates how position changes non-uniformly in discrete steps, and requires careful tracking of cumulative displacement — demonstrating that simple kinematic analysis extends to periodic back-and-forth motion.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

An object's position along the x-axis is given by x = a + bt² where a = 8.5 m and b = 2.5 m s⁻². Find the average velocity between t = 2 s and t = 4 s.

Position at t = 2 s: x(2) = 8.5 + 2.5(4) = 8.5 + 10 = 18.5 m. Position at t = 4 s: x(4) = 8.5 + 2.5(16) = 8.5 + 40 = 48.5 m.
Displacement = x(4) - x(2) = 48.5 - 18.5 = 30 m. Time interval = 4 - 2 = 2 s.
Average velocity = displacement / time = 30/2 = 15 m s⁻¹.
This is consistent with using v̄ = (v at t=2 + v at t=4)/2 = (5×2 + 5×4)/2 = (10+20)/2 = 15 m s⁻¹, since the acceleration is constant (a = 2b = 5 m s⁻²).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

Two trains A and B of length 400 m each are moving on parallel tracks at 72 km h⁻¹ and 36 km h⁻¹ respectively in the same direction. How long does train A take to pass train B completely?

Convert speeds: Train A: 72 km h⁻¹ = 20 m s⁻¹; Train B: 36 km h⁻¹ = 10 m s⁻¹.
Relative velocity of A with respect to B = vA - vB = 20 - 10 = 10 m s⁻¹.
Total length to be covered for A to completely pass B = length of A + length of B = 400 + 400 = 800 m.
Time = total length / relative velocity = 800/10 = 80 s. The concept of relative velocity reduces the two-train problem to an equivalent single-train problem with the relative speed.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 10

Applied Concept

A ball is dropped from a height of 90 m. Ignoring air resistance and taking g = 10 m s⁻², find its velocity just before hitting the ground and the time of fall.

Using v² = v0² + 2gh with v0 = 0 and h = 90 m: v² = 0 + 2(10)(90) = 1800, so v = √1800 ≈ 42.43 m s⁻¹.
Using v = v0 + gt: v = gt, so t = v/g = 42.43/10 ≈ 4.24 s. Alternatively, from h = ½gt²: t = √(2h/g) = √(2×90/10) = √18 ≈ 4.24 s.
The ball attains a speed of approximately 42.4 m s⁻¹ (about 153 km h⁻¹) after falling 90 m, illustrating how quickly speed builds up under gravity.
This is a straightforward free-fall application where kinematic equations with constant acceleration (g = 10 m s⁻²) directly yield both velocity and time.

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Chapter 2: Motion in a Straight Line | Application Questions

A car travelling at 126 km h⁻¹ is brought to rest within a distance of 200 m by uniform braking. Calculate the retardation of the car and the time taken to stop.

A player throws a ball vertically upward at 29.4 m s⁻¹. Using g = 9.8 m s⁻², find (a) the maximum height reached and (b) the time after which it returns to the player's hands.

A police van moving at 30 km h⁻¹ fires a bullet with muzzle speed 150 m s⁻¹ at a thief's car moving at 192 km h⁻¹ in the same direction. Find the speed of the bullet relative to the thief's car.

A woman walks from home to her office 2.5 km away at 5 km h⁻¹, stays all day, then returns by auto at 25 km h⁻¹. For the entire journey, find (a) average velocity and (b) average speed.

In a reaction time experiment, a ruler is dropped and falls 21 cm before being caught. Estimate the reaction time using g = 9.8 m s⁻².

With reference to a freely falling body starting from rest, show that the distances covered in the 1st, 2nd, and 3rd seconds are in the ratio 1:3:5.

A drunkard in a narrow lane takes 5 steps forward and 3 steps backward repeatedly, with each step being 1 m and taking 1 s. Determine after how many seconds he falls into a pit 13 m away.

An object's position along the x-axis is given by x = a + bt² where a = 8.5 m and b = 2.5 m s⁻². Find the average velocity between t = 2 s and t = 4 s.

Two trains A and B of length 400 m each are moving on parallel tracks at 72 km h⁻¹ and 36 km h⁻¹ respectively in the same direction. How long does train A take to pass train B completely?

A ball is dropped from a height of 90 m. Ignoring air resistance and taking g = 10 m s⁻², find its velocity just before hitting the ground and the time of fall.

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