Question

For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled and displacement of the particle?

Answer 1

Distance travelled = 42 m, Displacement = 22 m

Answer 2

• Understanding the graph: The given graph is a velocity-time (v-t) graph.

  • The area under the v-t graph represents displacement. Areas above the time axis are positive displacements, and areas below are negative displacements.
  • The total distance travelled is the sum of the magnitudes of all areas.

• Calculate areas for each segment:

  1. Segment 1 (0-2 s): Triangle above the t-axis. Area $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \text{ s} \times 8 \text{ m/s} = 8 \text{ m}$
  2. Segment 2 (2-4 s): Rectangle above the t-axis. Area $A_2 = \text{length} \times \text{width} = (4-2) \text{ s} \times 8 \text{ m/s} = 2 \text{ s} \times 8 \text{ m/s} = 16 \text{ m}$
  3. Segment 3 (4-6 s): Triangle above the t-axis. Area $A_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (6-4) \text{ s} \times 8 \text{ m/s} = \frac{1}{2} \times 2 \text{ s} \times 8 \text{ m/s} = 8 \text{ m}$
  4. Segment 4 (6-8 s): Triangle below the t-axis. Area $A_4 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (8-6) \text{ s} \times (-5) \text{ m/s} = \frac{1}{2} \times 2 \text{ s} \times (-5) \text{ m/s} = -5 \text{ m}$
  5. Segment 5 (8-10 s): Triangle below the t-axis. Area $A_5 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (10-8) \text{ s} \times (-5) \text{ m/s} = \frac{1}{2} \times 2 \text{ s} \times (-5) \text{ m/s} = -5 \text{ m}$

• Calculate Total Displacement: Displacement = Sum of all areas (with sign) $\Delta x = A_1 + A_2 + A_3 + A_4 + A_5 = 8 + 16 + 8 + (-5) + (-5) = 32 - 10 = 22 \text{ m}$

• Calculate Total Distance Travelled: Distance = Sum of magnitudes of all areas $D = |A_1| + |A_2| + |A_3| + |A_4| + |A_5| = 8 + 16 + 8 + |-5| + |-5| = 8 + 16 + 8 + 5 + 5 = 42 \text{ m}$