JEE PYQ: Motion in a Straight Line - Question ID 977dfda3305c (JEE Main 2021)

In problems involving motion with constant acceleration and deceleration, the following kinematic equations are fundamental:

• Velocity as a function of time: $v = u + at$
• Displacement as a function of time: $s = ut + \frac{1}{2} a t^2$
• Relation between velocity, acceleration, and displacement: $v^2 = u^2 + 2 a s$

Here, the car starts from rest ($u = 0$), accelerates at rate $\alpha$ for time $t_1$, then decelerates at rate $\beta$ for time $t_2$, coming to rest again. The total time is $t = t_1 + t_2$. The total distance travelled is the sum of the distances covered during acceleration and deceleration phases.

Step 1: Define the two phases of motion

Let $t_1$ be the time during which the car accelerates at $\alpha$, and $t_2$ be the time during which it decelerates at $\beta$. The total time is: $t = t_1 + t_2$

Step 2: Find the velocity at the end of acceleration phase

Since the car starts from rest, initial velocity $u = 0$. Using $v = u + a t$, the velocity at the end of acceleration phase is: $v = 0 + \alpha t_1 = \alpha t_1$ This velocity becomes the initial velocity for the deceleration phase.

Step 3: Use deceleration phase to find relation between $t_1$ and $t_2$

During deceleration, the car comes to rest, so final velocity $v_f = 0$. Using $v_f = v_i + a t$, where $v_i = \alpha t_1$ and $a = -\beta$: $0 = \alpha t_1 - \beta t_2 \implies \alpha t_1 = \beta t_2 \implies t_2 = \frac{\alpha}{\beta} t_1$

Step 4: Express total time in terms of $t_1$

Substitute $t_2$ into $t = t_1 + t_2$: $t = t_1 + \frac{\alpha}{\beta} t_1 = t_1 \left(1 + \frac{\alpha}{\beta}\right) = t_1 \frac{\alpha + \beta}{\beta} \implies t_1 = \frac{\beta}{\alpha + \beta} t$ Similarly, $t_2 = \frac{\alpha}{\beta} t_1 = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha + \beta} t = \frac{\alpha}{\alpha + \beta} t$

Step 5: Calculate distance travelled during acceleration phase

Using $s = ut + \frac{1}{2} a t^2$, with $u = 0$, $a = \alpha$, and $t = t_1$: $s_1 = 0 + \frac{1}{2} \alpha t_1^2 = \frac{1}{2} \alpha t_1^2$ Substitute $t_1$: $s_1 = \frac{1}{2} \alpha \left(\frac{\beta}{\alpha + \beta} t\right)^2 = \frac{1}{2} \alpha \frac{\beta^2}{(\alpha + \beta)^2} t^2$

Step 6: Calculate distance travelled during deceleration phase

During deceleration, initial velocity $u = \alpha t_1$, acceleration $a = -\beta$, time $t_2$. Using $s = ut + \frac{1}{2} a t^2$: $s_2 = \alpha t_1 \cdot t_2 + \frac{1}{2} (-\beta) t_2^2 = \alpha t_1 t_2 - \frac{1}{2} \beta t_2^2$ Substitute $t_1 = \frac{\beta}{\alpha + \beta} t$ and $t_2 = \frac{\alpha}{\alpha + \beta} t$: $$s_2 = \alpha \cdot \frac{\beta}{\alpha + \beta} t \cdot \frac{\alpha}{\alpha + \beta} t - \frac{1}{2} \beta \left(\frac{\alpha}{\alpha + \beta} t\right)^2 = \frac{\alpha \beta \alpha}{(\alpha + \beta)^2} t^2 - \frac{1}{2} \beta \frac{\alpha^2}{(\alpha + \beta)^2} t^2 = \frac{\alpha^2 \beta}{(\alpha + \beta)^2} t^2 - \frac{\alpha^2 \beta}{2(\alpha + \beta)^2} t^2 = \frac{\alpha^2 \beta}{2(\alpha + \beta)^2} t^2$$

Step 7: Total distance travelled

Total distance $S = s_1 + s_2$: $$S = \frac{1}{2} \alpha \frac{\beta^2}{(\alpha + \beta)^2} t^2 + \frac{\alpha^2 \beta}{2(\alpha + \beta)^2} t^2 = \frac{\alpha \beta^2 + \alpha^2 \beta}{2(\alpha + \beta)^2} t^2 = \frac{\alpha \beta (\beta + \alpha)}{2(\alpha + \beta)^2} t^2 = \frac{\alpha \beta (\alpha + \beta)}{2(\alpha + \beta)^2} t^2 = \frac{\alpha \beta}{2(\alpha + \beta)} t^2$$

Conclusion:

The total distance travelled is: $\boxed{ \frac{\alpha \beta}{2(\alpha + \beta)} t^2 }$ This matches option C.

Students often make the following mistakes:

• Incorrectly assuming equal time intervals: Many assume $t_1 = t_2 = t/2$, which is only true if $\alpha = \beta$. This leads to wrong expressions.
• Sign errors in deceleration: Forgetting that deceleration is negative acceleration and misapplying the kinematic equations.
• Algebraic simplification errors: Especially when combining terms like $\alpha \beta^2 + \alpha^2 \beta$, students may factor incorrectly.
• Using average velocity incorrectly: While average velocity can be used, it must be applied carefully over each phase, not over the entire motion.

Exam Tip: Always define variables clearly, break the motion into phases, and verify units and dimensions at each step. The final expression must have dimensions of length, so $\frac{\alpha \beta}{(\alpha + \beta)} t^2$ has correct dimensions (since $\alpha t^2$ and $\beta t^2$ are lengths).