# Answer to Question #11477 in Mechanics | Relativity for jo

Question #11477

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 second evaluate a)the max. velocity b)the total distance travelled

Expert's answer

Let's denote Ta the acceleration time of a car and Td the deceleration time. Then

a·Ta = |b|·Td

Also we know that

Ta + Td = T.

So,

Td = T - Ta = a·Ta/|b| ==> Ta(a/|b|+1) = T ==> Ta = T/(a/|b|+1),

and the maximum velocity of a car is

Vmax = a·T/(a/|b|+1).

The total travelled distance is

D = Da + Dd = a·Ta²/2 + b·Td²/2 = a·Ta²/2 + b·(T-Ta)²/2 = a·(T/(a/|b|+1))²/2 + b·(T-T/(a/|b|+1))²/2

a·Ta = |b|·Td

Also we know that

Ta + Td = T.

So,

Td = T - Ta = a·Ta/|b| ==> Ta(a/|b|+1) = T ==> Ta = T/(a/|b|+1),

and the maximum velocity of a car is

Vmax = a·T/(a/|b|+1).

The total travelled distance is

D = Da + Dd = a·Ta²/2 + b·Td²/2 = a·Ta²/2 + b·(T-Ta)²/2 = a·(T/(a/|b|+1))²/2 + b·(T-T/(a/|b|+1))²/2

## Comments

## Leave a comment