NEET 2024 Physics Question 41 Solution

The instantaneous acceleration a(t) is the first time derivative of velocity, which corresponds graphically to the slope of the tangent to the velocity-time graph at that instant:

Analyzing the given trapezoidal v-t curve: 1. $0 \le t < t_1$: $v = 0 \implies a = 0$. 2. $t_1 \le t < t_2$: $v$ is an upward inclined straight line with constant positive slope: $$a = \frac{dv}{dt} = +a_0 = \text{constant} > 0$$ 3. $t_2 \le t < t_3$: $v$ is a horizontal plateau with zero slope: $$a = \frac{dv}{dt} = 0$$ 4. $t_3 \le t < t_4$: $v$ is a downward inclined straight line with constant negative slope: $$a = \frac{dv}{dt} = -a_0 = \text{constant} < 0$$ 5. $t \ge t_4$: $v = 0 \implies a = 0$.

The corresponding a-t graph consists of a positive rectangular pulse, followed by a flat zero line, followed by an equal negative rectangular pulse below the time axis. This uniquely matches Graph (1).