A car acceleration form 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 elapsed is t evaluate (a) the maximum velocity attained and (b) the total distance travelled.

Options

A. `(ab)/(a+b)tm//s`
B. `(ab)/(a-b)t m//s`
C. `(2ab)/(a+b)t m//s`
D. `(2ab)/(a-b)t m//s`

Answer

A

Explanation

Total time of motion =t.
Duration of acceleration=t'
Duration of deceleration `=t=t`'.
Given u=0, a = constant acceleration and b=constant deceleration.
`v=0+at`'............(i)
Also `0=v-b(t-1')` .............(ii)
From (ii) `-v=-bt+bt'rArr-at'=-bt+bt'`
`rArr(a+b)t'rArrt'=(b)/((a+b))t`.
But v=at'.
`therefore` Maximum velocity attained=at'
`rArrv=(ab)/((a+b))t=m//s`.