A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a constan

Question

A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$, then the maximum velocity acquired by the car is:

Accepted Answer

1. **Define Variables:** Let the maximum velocity acquired be $v_{max}$. Let the time taken to accelerate be $t_1$ and the time taken to decelerate be $t_2$. The total time is $t = t_1 + t_2$.
2. **Analyze Acceleration Phase:** The car starts from rest ($u=0$) and accelerates at $\alpha$. Using the first equation of motion ($v = u + at$) :
   $$ v_{max} = 0 + \alpha t_1 \implies t_1 = \frac{v_{max}}{\alpha} $$
3. **Analyze Deceleration Phase:** The car starts from $v_{max}$ and decelerates at $\beta$ to rest ($v=0$).
   $$ 0 = v_{max} - \beta t_2 \implies v_{max} = \beta t_2 \implies t_2 = \frac{v_{max}}{\beta} $$
4. **Combine and Solve:** Substitute $t_1$ and $t_2$ into the total time equation:
   $$ t = t_1 + t_2 = \frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta} $$
   $$ t = v_{max} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = v_{max} \left( \frac{\alpha + \beta}{\alpha\beta} \right) $$
   $$ v_{max} = \frac{\alpha\beta t}{\alpha + \beta} $$

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