A car starts from rest and travels with uniform acceleration a, for some time and then with uniform retardation `beta` and comes to rest. If the total time of car is 't', the maximum velocity attained by it is given by

Explanation

Acceleration `a=alpham//sec^(2)`
Initial speed `u=0m//sec`
Let the time be `t_(1)` sec
From equation `v=u+atrArrv=0+alphat_(1)`
`thereforet_(1)=v/alphasec`
Retardation `a=-betam//sec^(2)`
Initial speed 'u' is equal to the final velocity with acceleration `'alpha'`
= `u = alphat_(1)m//sec`
Final velocity v = 0 m/sec
Let the time be `t_(2)` sec.
From equation, v = u + at
`rArr0=alphat_(1)+(-B)t_(2)`
`thereforealphat_(1)=betat_(2)" "(becausev=alphat_(1))`
`v=betat_(2)rArrt_(2)=v/beta`
Total time of motion `t=t_(1)+t_(2)`
`t=v/alpha+v/betarArrt=v(1/alpha+1/beta)`
`rArrt=v((alpha+beta)/(alphabeta))rArrv=(alphabetat)/(alpha+beta)`
`therefore` The maximum velocity attained by the car `=(alphabetat)/(alpha+beta)`