A car accelerates from rest at a constant rate a for sometime after which it
decelerates at a constant rate B to come to rest. If the total time elapsed is t sec, evaluate
(a) the maximum velocity reached.
(b) the total distance traversed.
the its
Answers
Answered by
5
Answer:
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)
Hope you like this. Mark me as brainliest
Explanation:
Similar questions