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 , evauate.
(i) maximum velocity reached , and
(ii) the total distance travelled .

Explanation

If the car has uniform acceleration for time (t) and uniform retardation for time ` t-2`, then total time is ` T= t_1 + t_2` …(i)`
.
If ` v_(max)` be the maximum velocity of the car. As slope of ` v-t` graph give acceleration, so
` A= ( v_max)/t-1` and ` B = ( v_(max)/t-2`
Thus, ` 1/ A + 1 / B = ( t_1 + t_2)/(v_(max) [from (i) ]`
or ` v-(max) = (ABT)/(A+B)` ...(i)
we know that area under ` v-t` graph gives the distance travelled in a given time. So`
` S= 1/2 (v_(max) xx T= ( ABT)^2/(2 ( A+B))`.