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

Options

A. `(alphabeta)/(alpha+beta)t`
B. `1/2(alpha beta)/(alpha+beta)t^2`
C. `(alphabeta)/(alpha-beta)t`
D. `1/2(alphabeta)/(alpha-beta)t^2`

Explanation

Let,
`barP=6hati+6hatj-3hatk`
`barQ=7hati+4hatj-4hatk`
Angle between `barP`and `barQ` is,`costheta=(barP.barQ)/(|barP||barQ|)`
`barP.barQ=(6hati+6hatj-3hatk).(7hati+4hatj-4hatk)`
`=42+24-12=54`
`|barP|=sqrt(6^2+6^2+3^2)=9`
`|barQ|=sqrt(7^2+4^2+4^2)=9`
`impliescostheta=54/(9xx9)`
`theta=cos^-1((2)/3)" or "sin^1(sqrt(5)/3)`