JEE Advance - Physics (1978 - No. 2)
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.
Maximum velocity: $$\frac{\alpha \beta}{\alpha + \beta}t$$, Total distance: $$\frac{1}{2} \frac{\alpha \beta}{\alpha + \beta} t^2$$
Maximum velocity: $$\frac{\alpha + \beta}{\alpha \beta}t$$, Total distance: $$2 \frac{\alpha + \beta}{\alpha \beta} t^2$$
Maximum velocity: $$\frac{\alpha}{\beta}t$$, Total distance: $$\frac{1}{2} \frac{\alpha}{\beta} t^2$$
Maximum velocity: $$\alpha \beta t$$, Total distance: $$\frac{1}{2} \alpha \beta t^2$$
Maximum velocity: $$\frac{\beta}{\alpha}t$$, Total distance: $$\frac{1}{2} \frac{\beta}{\alpha} t^2$$
Comments (0)
