The velocity (v) - Distance (x) graph is shown in figure. Which graph represents acceleration(a) versus distance ( x ) variation of this system?

Solution

Physics sir · 4 steps

This is velocity against distance, not time, so use a = vdv/dx, not the slope alone.

1. Read the graph. The line starts at v₀ when x = 0 and falls steadily with slope -k. So v = v₀ - kx, and v becomes zero at x = v₀/k.
2. Acceleration from v-x. Acceleration is dv/dt = dv/dx·dx/dt = vdv/dx. Here dv/dx = -k, a constant.

   a = (v₀ - kx)(-k) = -kv₀ + k² x

3. Shape of a-x. This is a straight line with a negative intercept -kv₀ and a positive slope k². It crosses a = 0 at x = v₀/k, exactly where v becomes zero.
4. So. The a-x graph starts below the axis and rises through it, which is option (d).

The idea to keep. Whenever velocity is plotted against position, acceleration is vdv/dx. A straight v-x line gives a straight a-x line, not a constant acceleration.

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