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Question: Velocity-time (v-t) graph for a moving object is shown in the figure. Total displacement of the obje...

Velocity-time (v-t) graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is

A

60 m

B

50 m

C

30 m

D

40 m

Answer

50 m

Explanation

Solution

The displacement of an object from a velocity-time (v-t) graph is given by the area under the graph. Acceleration is given by the slope of the v-t graph.

  • A positive slope indicates acceleration.
  • A negative slope indicates retardation (deceleration).
  • A zero slope (horizontal line) indicates zero acceleration (constant velocity).

We need to find the total displacement during the time intervals when there is non-zero acceleration and retardation.

Let's analyze the graph segments:

  1. From t = 0 s to t = 20 s: The velocity is constant at 1 m/s. The slope is zero, so acceleration is zero. This interval is not included.

  2. From t = 20 s to t = 30 s: The velocity increases linearly from 1 m/s to 4 m/s. The slope is positive, indicating non-zero acceleration. This segment forms a trapezium.

    Displacement (S1) = Area of trapezium = 12×(sum of parallel sides)×height\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

    S1=12×(1 m/s+4 m/s)×(30 s−20 s)S_1 = \frac{1}{2} \times (1 \text{ m/s} + 4 \text{ m/s}) \times (30 \text{ s} - 20 \text{ s})

    S1=12×(5 m/s)×(10 s)S_1 = \frac{1}{2} \times (5 \text{ m/s}) \times (10 \text{ s})

    S1=25 mS_1 = 25 \text{ m}

  3. From t = 30 s to t = 40 s: The velocity decreases linearly from 4 m/s to 1 m/s. The slope is negative, indicating non-zero retardation. This segment also forms a trapezium.

    Displacement (S2) = Area of trapezium = 12×(sum of parallel sides)×height\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

    S2=12×(4 m/s+1 m/s)×(40 s−30 s)S_2 = \frac{1}{2} \times (4 \text{ m/s} + 1 \text{ m/s}) \times (40 \text{ s} - 30 \text{ s})

    S2=12×(5 m/s)×(10 s)S_2 = \frac{1}{2} \times (5 \text{ m/s}) \times (10 \text{ s})

    S2=25 mS_2 = 25 \text{ m}

  4. From t = 40 s to t = 60 s: The velocity is constant at 1 m/s. The slope is zero, so acceleration is zero. This interval is not included.

The total displacement of the object during the time interval when there is non-zero acceleration and retardation is the sum of the displacements from 20 s to 30 s and from 30 s to 40 s.

Total Displacement = S1+S2=25 m+25 m=50 mS_1 + S_2 = 25 \text{ m} + 25 \text{ m} = 50 \text{ m}