A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time lapse is $t$ seconds, evaluate. (i) maximum velocity reached, and (ii) the total distance travelled.

Step 1: Define the time intervals for acceleration and deceleration.

Let $t_1$ be the acceleration time and $t_2$ be the deceleration time.

$$t_1 + t_2 = t$$

Step 2: Express the maximum velocity in terms of acceleration and deceleration.

The maximum velocity $v_{\max}$ is reached at the end of the acceleration phase.

$$v_{\max} = \alpha t_1$$ $$v_{\max} = \beta t_2$$

Step 3: Solve for the time intervals $t_1$ and $t_2$.

Equate the two expressions for $v_{\max}$ and substitute $t_2 = t - t_1$.

$$\alpha t_1 = \beta (t - t_1)$$ $$t_1 = \frac{\beta t}{\alpha + \beta}$$ $$t_2 = \frac{\alpha t}{\alpha + \beta}$$

Step 4: Calculate the maximum velocity.

Substitute the expression for $t_1$ into the equation for $v_{\max}$.

$$v_{\max} = \alpha \left( \frac{\beta t}{\alpha + \beta} \right)$$ $$v_{\max} = \frac{\alpha \beta t}{\alpha + \beta}$$

Step 5: Calculate the total distance travelled.

The total distance $s$ is the area under the triangular velocity-time graph.

$$s = \frac{1}{2} t v_{\max}$$ $$s = \frac{1}{2} t \left( \frac{\alpha \beta t}{\alpha + \beta} \right)$$ $$s = \frac{\alpha \beta t^2}{2(\alpha + \beta)}$$

Answer: (i) Maximum velocity reached: $v_{\max} = \dfrac{\alpha\beta t}{\alpha+\beta}$ (ii) Total distance travelled: $s = \dfrac{\alpha\beta t^2}{2(\alpha+\beta)}$