Question

A particle starts from rest and accelerate at rate a
over a distance x and then retards to rest at rate
b over a distance y. Then the maximum velocity of
particle is what?​

Answer 1

Answer:The maximum velocity would be ab(T)/ a+b assuming that 'T' is the total time taken by the particle.

Explanation:

Let total time of motion of particle be T

The particle accelerates till time t1 at a rate a

The particle decelerates or retards during time (T-t1) at a rate b

Let maximum velocity be V

From velocity-time graph plotted(as shown in the image attached)

V/t1 = a ie, V= at1 ...............(1)

V/(T-t1) = b ie, V= b(T-t1) ..........(2)

On equating (1) & (2)

at1= b(T-t1)

t1 = bT/a+b.........(3)

Substituting (3) in (1),

V=at1= abT/a+b