Question

a car accelerates frm rest at a constant rate 'A' for
some time after which it decelerate at a constant rate
'B' and comes to rest .if total tym elapsed is 't' then
maximum velocity acquired by car will be.?

Answer 1

Answer on Question #43335, Physics, Mechanics | Kinematics | Dynamics

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

Solution:

For the first period of motion the acceleration is

$$a_1 = A = \frac{v - v_0}{t_1} = \frac{v}{t_1}$$ $$t_1 = \frac{v}{A}$$

For the second period of motion the acceleration is

$$a_2 = -B = \frac{0 - v}{t_2} = -\frac{v}{t_2}$$ $$t_2 = \frac{v}{B}$$

From given

$$t = t_1 + t_2 = \frac{v}{A} + \frac{v}{B} = v \left(\frac{1}{A} + \frac{1}{B}\right) = v \frac{A + B}{AB}$$

Thus,

$$v = \frac{AB}{A + B} t$$

Answer: $v = \frac{AB}{A + B} t$

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