JEE Main 2021 · 31 August · evening · Numerical answer · +4 / −0

A particle is moving with constant acceleration 'a'. Following graph shows v² versus x(displacement) plot. The acceleration of the particle is ___________ m/s².

JEE Main PYQ figure
Answer and solution

Answer 1

Solution

Physics sir · 3 steps

A straight line on a v² versus x graph means its slope is 2a, so read the slope and halve it.

  1. Pick the equation. For constant acceleration, v² = u² + 2ax. Plotting v² against x gives a straight line with intercept u² and slope 2a.

    slope = (Δ (v²))/(Δ x) = 2a

  2. Read the slope. From point A (10, 40) to point C (30, 80): slope = (80 - 40)/(30 - 10) = 40/20 = 2 m s⁻². The intercept at x = 0 is 20, so u² = 20 m² s⁻², which fits the same line.
  3. So. 2a = 2, so a = 1 m s⁻².

The common slip: 2. Taking the slope itself as the acceleration gives 2, not 1. The slope of this graph is 2a.

The idea to keep. On a v² versus x graph, the slope is 2a and the intercept is u². Always halve the slope to get the acceleration.

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 2021.