JEE Main 2023 · 24 January · evening · Single correct · +4 / −1

The velocity time graph of a body moving in a straight line is shown in the figure.
The ratio of displacement to distance travelled by the body in time 0 to 10s is :

JEE Main PYQ figure
  1. (a) 1 : 1
  2. (b) 1 : 3
  3. (c) 1 : 4
  4. (d) 1 : 2
Answer and solution

Answer (b) 1 : 3

Solution

Physics sir · 3 steps

Area under the v-t graph: keep the signs for displacement, drop them for distance.

  1. Read the areas. On a velocity-time graph, the area between the curve and the time axis gives the displacement in that interval. Area above the axis is positive and area below it is negative. Each piece here is a rectangle, so area is velocity × time.
    IntervalVelocitySigned areaSize of area
    0 to 2 s8 m s⁻¹+16 m16 m
    2 to 4 s-4 m s⁻¹-8 m8 m
    4 to 8 s4 m s⁻¹+16 m16 m
    8 to 10 s-4 m s⁻¹-8 m8 m
  2. Add them up. Displacement = 16 - 8 + 16 - 8 = 16 m. Distance = 16 + 8 + 16 + 8 = 48 m.
  3. So. Displacement : distance = 16 : 48 = 1 : 3, which is option (b).

The idea to keep. Displacement is the signed area under a v-t graph and distance is the total area with every part counted as positive. They are equal only if the body never reverses direction.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Rectilinear Motion

Asked about 110 times in JEE Main since 2020. This question is from 2023.