Question

a car accelerates from rest at a constant rate alpha for some time, after which it decelerates at a constant rate beeta and comes to rest. if the total time elapsed is t, the maximum velocity acquired by the car will be,t distance covered by car in term of alpha,bita and t

Answer 1

alpha×t1=Vmax
beta(T-t1)=Vmax
alphat1=beta(T-t1)
alphat1=betaT-betat1
alphat1+betat1=betaT
(alpha+beta)t1=betaT
t1=betaT/alpha+beta
Vmax=alphat1
=alphabetaT/(alpha+beta)

S=1/2×Vmax×T
=1/2×alphabetaT/(alpha+beta)×T
=alphabetaT^2/2(alpha+beta)