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

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

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Case Set 1

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A traffic police officer is studying road safety at a busy highway junction. He observes that vehicles travelling at high speeds take much longer to stop after brakes are applied. He records that a particular car moving at 20 m s⁻¹ is brought to rest within a distance of 40 m when brakes are applied uniformly. The officer also notes that the driver's reaction time before applying the brakes is approximately 0.2 s. The officer uses this data to recommend a safe speed limit for the stretch of road, keeping in mind that pedestrians may suddenly appear and the total stopping distance must not exceed 50 m.

Question 1: Calculate the deceleration of the car after brakes are applied.

Using the kinematic equation v² = v0² + 2ax, with v = 0, v0 = 20 m s⁻¹, and x = 40 m: 0 = 400 + 2a(40).
Solving: a = -400/80 = -5 m s⁻²; the magnitude of deceleration is 5 m s⁻².
This deceleration is caused by braking force and represents uniform retardation applied after the brakes engage.

Question 2: Find the distance covered by the car during the driver's reaction time.

During reaction time (tr = 0.2 s), the car continues at its initial speed of 20 m s⁻¹ before brakes are applied.
Reaction distance = v0 × tr = 20 × 0.2 = 4 m.
Thus the car travels 4 m before deceleration begins, which must be added to braking distance for total stopping distance.

Question 3: The officer wants to recommend a speed limit such that total stopping distance (reaction distance + braking distance) does not exceed 50 m, assuming the same deceleration of 5 m s⁻² and reaction time of 0.2 s. Find the maximum safe speed.

Total stopping distance = v0 × tr + v0²/(2a) ≤ 50 m. With tr = 0.2 s and a = 5 m s⁻²: 0.2v0 + v0²/10 ≤ 50.
Let v0 = x: x²/10 + 0.2x - 50 ≤ 0, i.e., x² + 2x - 500 ≤ 0. Solving x² + 2x - 500 = 0: x = (-2 + √2004)/2 ≈ (-2 + 44.78)/2 ≈ 21.4 m s⁻¹.
The safe speed limit is approximately 21.4 m s⁻¹ ≈ 77 km h⁻¹; speeds above this would require more than 50 m to stop, making the stretch unsafe.

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Case Set 2

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During a school science fair, students conduct an experiment to measure human reaction time. Each student is asked to catch a ruler dropped vertically by a partner without warning. The ruler is held at the top and released; the catcher's fingers are positioned at the zero mark. After catching, the student reads the distance d the ruler has fallen. Student A catches it at d = 21 cm, student B at d = 11.25 cm, and student C at d = 30 cm. The teacher explains that the ruler undergoes free fall from rest (neglecting air resistance) and that g = 9.8 m s⁻² can be used to convert the fall distance directly into reaction time. The results are used to compare reaction times among students.

Question 1: State the formula used to calculate reaction time from the distance d fallen by the ruler.

The ruler falls from rest under free fall: d = ½g·tr², where tr is the reaction time.
Rearranging: tr = √(2d/g), which directly gives reaction time from the measured distance.
This formula comes from the second kinematic equation with initial velocity zero and acceleration g.

Question 2: Calculate the reaction time of student A (d = 21 cm) and student B (d = 11.25 cm).

For student A: tr = √(2 × 0.21/9.8) = √(0.04286) ≈ 0.207 s ≈ 0.2 s.
For student B: tr = √(2 × 0.1125/9.8) = √(0.02296) ≈ 0.1515 s ≈ 0.15 s.
Student B has a shorter reaction time, indicating faster response; the difference illustrates individual variation in reaction time as noted in the chapter.

Question 3: Student C (d = 30 cm) was distracted during the experiment. Compare her reaction time with student A and explain how reaction time affects the stopping distance of a car moving at 15 m s⁻¹ with braking deceleration of 7.5 m s⁻².

Reaction time of C: tr = √(2 × 0.30/9.8) = √(0.06122) ≈ 0.247 s ≈ 0.25 s; student A's tr ≈ 0.2 s. Student C is slower by ≈ 0.05 s.
Reaction distance for A: 15 × 0.2 = 3 m; braking distance = 15²/(2 × 7.5) = 225/15 = 15 m; total = 18 m.
Reaction distance for C: 15 × 0.25 = 3.75 m; total stopping distance = 3.75 + 15 = 18.75 m. Even a 0.05 s slower reaction adds 0.75 m extra stopping distance — significant in emergency braking scenarios.

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Case Set 3

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A physics teacher demonstrates motion graphs using a battery-powered toy car on a straight track. She programs the car to move as follows: it starts from rest at the origin, accelerates uniformly to 4 m s⁻¹ in 4 seconds, then moves at constant 4 m s⁻¹ for 4 more seconds, then decelerates uniformly back to rest in the next 4 seconds. Students are asked to sketch and analyse the x-t, v-t, and a-t graphs for the entire 12-second journey. The teacher uses this demonstration to help students understand the relationships between position, velocity, and acceleration, and the significance of slope and area in motion graphs.

Question 1: Describe the shape of the v-t graph for the toy car and identify the acceleration during each phase.

Phase 1 (0-4 s): v-t graph is a straight line rising from 0 to 4 m s⁻¹; slope = acceleration = (4-0)/4 = 1 m s⁻².
Phase 2 (4-8 s): v-t graph is a horizontal line at v = 4 m s⁻¹; slope = 0, so acceleration = 0 m s⁻².
Phase 3 (8-12 s): v-t graph is a straight line falling from 4 m s⁻¹ to 0; slope = (0-4)/4 = -1 m s⁻² (deceleration).

Question 2: Calculate the total displacement of the toy car over the entire 12-second journey using the area under the v-t graph.

Area under v-t graph equals displacement. Phase 1: area = ½ × 4 × 4 = 8 m (triangle).
Phase 2: area = 4 × 4 = 16 m (rectangle). Phase 3: area = ½ × 4 × 4 = 8 m (triangle).
Total displacement = 8 + 16 + 8 = 32 m. This matches using kinematic equations for each phase separately, confirming the area-displacement relationship.

Question 3: Sketch the nature of the x-t graph for all three phases and explain what features of the curve correspond to each phase of motion.

Phase 1 (0-4 s): x-t graph is a parabola curving upward (concave up) starting from the origin, since x = ½(1)t² — positive acceleration means increasing slope.
Phase 2 (4-8 s): x-t graph is a straight line with constant positive slope (v = 4 m s⁻¹), since zero acceleration means constant velocity — the line continues from x = 8 m.
Phase 3 (8-12 s): x-t graph is a parabola curving downward (concave down), since negative acceleration means decreasing slope; the curve levels off to a horizontal tangent at t = 12 s when the car stops at x = 32 m.

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Case Set 4

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Two friends Arjun and Priya live on the same straight road. Arjun's house is taken as the origin. Priya lives 500 m away in the positive direction. One morning, Arjun starts cycling toward Priya's house at 5 m s⁻¹. At the same moment, Priya starts walking toward Arjun's house at 1.5 m s⁻¹. Both move with constant velocities. They wish to find where and when they will meet. Later, Arjun turns around immediately after reaching Priya's house and cycles back at the same speed of 5 m s⁻¹. They use the concept of relative velocity to simplify their calculations.

Question 1: Using relative velocity, find the time after which Arjun and Priya meet for the first time.

Arjun's velocity = +5 m s⁻¹ (toward Priya); Priya's velocity = -1.5 m s⁻¹ (toward Arjun). Both move toward each other, so relative velocity of approach = 5 + 1.5 = 6.5 m s⁻¹.
Initial separation = 500 m. Time to meet = 500/6.5 ≈ 76.9 s.
At this time, Arjun has travelled 5 × 76.9 = 384.6 m from his house — the meeting point is ≈ 384.6 m from the origin.

Question 2: What is the velocity of Arjun relative to Priya, and of Priya relative to Arjun, when they move toward each other?

Taking positive direction as from Arjun toward Priya: vArjun = +5 m s⁻¹, vPriya = -1.5 m s⁻¹.
Velocity of Arjun relative to Priya = vArjun - vPriya = 5 - (-1.5) = +6.5 m s⁻¹ (Arjun approaches Priya at 6.5 m s⁻¹).
Velocity of Priya relative to Arjun = vPriya - vArjun = -1.5 - 5 = -6.5 m s⁻¹ (equal magnitude, opposite direction — confirming vAB = -vBA).

Question 3: After meeting Priya, Arjun continues to her house (500 m from origin), immediately turns back at 5 m s⁻¹, and Priya continues toward the origin at 1.5 m s⁻¹. Find the time elapsed from the start until Arjun and Priya meet for the second time.

First meeting at t1 ≈ 76.9 s: Arjun at x ≈ 384.6 m, Priya at x ≈ 384.6 m. Arjun still needs 500 - 384.6 = 115.4 m to reach Priya's house; time = 115.4/5 ≈ 23.1 s, so Arjun reaches Priya's house at t ≈ 100 s.
At t = 100 s, Priya's position = 500 - 1.5 × 100 = 500 - 150 = 350 m. Arjun turns back (velocity = -5 m s⁻¹); separation = 500 - 350 = 150 m.
Relative velocity of approach (now same direction analysis): Arjun moves at -5 m s⁻¹ toward origin, Priya at -1.5 m s⁻¹ toward origin; Arjun gains on Priya at 5 - 1.5 = 3.5 m s⁻¹. Time to close 150 m = 150/3.5 ≈ 42.9 s. Second meeting at t ≈ 100 + 42.9 ≈ 142.9 s from start.

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Case Set 5

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A ball is thrown vertically upward from the roof of a building 25 m high with an initial velocity of 15 m s⁻¹. Taking the ground as the origin and upward as the positive direction, and using g = 10 m s⁻², students in a physics class are asked to analyse the complete motion of the ball — from the moment it is thrown until it strikes the ground. The teacher points out that using a single kinematic equation (second method) is more efficient than splitting the path into upward and downward segments. Students are also asked to interpret the sign of velocity and acceleration at various stages of the journey.

Question 1: State the values of initial position y0, initial velocity v0, and acceleration a to be substituted in the kinematic equations for this problem.

Initial position: y0 = 25 m (the roof is 25 m above the ground, which is the origin).
Initial velocity: v0 = +15 m s⁻¹ (positive because the ball is thrown upward, which is the positive direction).
Acceleration: a = -g = -10 m s⁻² (gravity acts downward, opposite to the chosen positive direction).

Question 2: Find the maximum height reached by the ball above the ground.

At maximum height, v = 0. Using v² = v0² + 2a(y - y0): 0 = (15)² + 2(-10)(y - 25).
Solving: 20(y - 25) = 225, so y - 25 = 11.25 m, giving y = 36.25 m above the ground.
The ball rises 11.25 m above the roof, reaching a maximum height of 36.25 m above the ground.

Question 3: Using the single-equation method, find the total time the ball takes to reach the ground after being thrown. Explain why this method is preferred over the two-stage method.

Using y = y0 + v0t + ½at²: 0 = 25 + 15t + ½(-10)t² = 25 + 15t - 5t². Rearranging: 5t² - 15t - 25 = 0, or t² - 3t - 5 = 0.
Solving: t = [3 ± √(9 + 20)]/2 = [3 ± √29]/2 = [3 ± 5.385]/2. Taking positive root: t = (3 + 5.385)/2 ≈ 4.19 s.
The single-equation method is preferred because it handles the entire journey (upward and downward) in one step without needing to track path segments separately, reducing the risk of sign errors and saving calculation time.

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Chapter sections

Chapter 2: Motion in a Straight Line | Case Study Questions

Passage

Question 1: Calculate the deceleration of the car after brakes are applied.

Question 2: Find the distance covered by the car during the driver's reaction time.

Question 3: The officer wants to recommend a speed limit such that total stopping distance (reaction distance + braking distance) does not exceed 50 m, assuming the same deceleration of 5 m s⁻² and reaction time of 0.2 s. Find the maximum safe speed.

Passage

Question 1: State the formula used to calculate reaction time from the distance d fallen by the ruler.

Question 2: Calculate the reaction time of student A (d = 21 cm) and student B (d = 11.25 cm).

Question 3: Student C (d = 30 cm) was distracted during the experiment. Compare her reaction time with student A and explain how reaction time affects the stopping distance of a car moving at 15 m s⁻¹ with braking deceleration of 7.5 m s⁻².

Passage

Question 1: Describe the shape of the v-t graph for the toy car and identify the acceleration during each phase.

Question 2: Calculate the total displacement of the toy car over the entire 12-second journey using the area under the v-t graph.

Question 3: Sketch the nature of the x-t graph for all three phases and explain what features of the curve correspond to each phase of motion.

Passage

Question 1: Using relative velocity, find the time after which Arjun and Priya meet for the first time.

Question 2: What is the velocity of Arjun relative to Priya, and of Priya relative to Arjun, when they move toward each other?

Question 3: After meeting Priya, Arjun continues to her house (500 m from origin), immediately turns back at 5 m s⁻¹, and Priya continues toward the origin at 1.5 m s⁻¹. Find the time elapsed from the start until Arjun and Priya meet for the second time.

Passage

Question 1: State the values of initial position y0, initial velocity v0, and acceleration a to be substituted in the kinematic equations for this problem.

Question 2: Find the maximum height reached by the ball above the ground.

Question 3: Using the single-equation method, find the total time the ball takes to reach the ground after being thrown. Explain why this method is preferred over the two-stage method.

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