Question

A particle starts from rest with constant acceleration a and it is then decelerated with a constant value b
till it is brought to rest. If the total time taken between these two rest positions ist. What is the maximum
speed acquired by the particle?​

Answer 1

Answer:

The motion can be divided into two parts

1. When accelaration is equal to a.

2. When acceleration is equal to b.

During first motion let v be its final speed and x be its time taken

Then

V=u+at

V=0+ax

X=v÷a (eq 1)

And in second motion v is its initial speed and final speed is 0 and let y be time taken.

Therefore

V=u+at

0=v-by because acc is negative

Y=v÷b (eq 2)

By adding eq 1 and eq 2

X+y= v(a+b)÷ab

T=v(a+b)÷ab because total time is given as T

V=abT÷(a+b)

Explanation:

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