A car accelerates from rest at a constant rate alpha for sometime after which it decelerates at a constant rate beta to come to rest. If the total tiem lapsed in t seconds , find (1)the maximum velocity reached and (2)the total distance travelled.

A car accelerates from rest at a constant rate alpha for sometime afte

1.	A car accelerates from rest at a constant rate alpha for sometime after which it decelerates at a constant rate beta to come to rest. If the total tiem lapsed in t seconds , find (1)the maximum velocity reached and (2)the total distance travelled.
Answer» Solution :LET` v_(m)` be the maximum velocity and let `t_(1)` be time taken to attain it . Then using `v=u+at` we get `v_(m)=ALPHA t_(1)`……(1) Let `t_(2)` be the time taken by the car to stop under retardation `BETA`. Then `0=v_(m) beta t_(2)"or"v_(m)beta t_(2)`....(2) Eqns.(1)and(2) GIVEN `t_(2)/t_(1)=alpha/beta"or" t_(2)/t_(1)+1-alpha/beta+1"or" t/t_(1)=(alpha+beta)/beta"or" t_(1)=beta/(alpha+beta)t` Substituting this value of `t_(1)`in eqn.(1), `v_(m)=(alphabeta)/(alpha+beta)t`t Now , let `s_(1)`be the distance travelled during acceleration and let `s_(2)`be the distance travelled during retardation . Then using the equation `v^(2)=u^(2)+2as`,we get `v_(m)^(2)=2alphas_(1)`........(3) and `0=v_(m)^(2)-2betas_(2)`.......(4) `:. s_(1)=v_(m)^(2)/(2alpha)=(alphabeta^(2))/(2(alpha+beta)^(2)t^(2)"and"s_(2)` `=(alpha^(2)beta)/(2(alpha+beta))^(2)t^(2)` `:. `Total distance `s=s_(1)+s_(2)` `=(alphabetat^(2))/(2(alpha+beta))^(2)(alpha+beta)=(alphabetat^(2))/(2(alpha+beta))`