Question

A car accelerates from rest at a constant rate 'a' for some time after which it decelerates at a constant rate 'B' and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be

Answer 1

a car starts moving from rest : let initial velocity u=0 m/sec
it gains max velocity in time  t1.  
the acceleration of the car during this period t1 be 'A'
then  v=u+at
=>  umax= 0 + At1
=>  umax = At1
=>  t1=umax/A ------------------------------- 1

now the car decelerates  for time t2 and comes to rest.
in this case u= umax, a=-B, time t=t2, final velocity v=0;
 
v=u+at
=> 0= umax + (-B)t2
=> Bt2=umax.
=>t2=umax/B

now total time taken is T=t1+t2
T=umax/A + umax/B
=> T=umax(A+B)/AB
=> umax=T.AB/(A+B)

 
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