Question

The initial velocity of a particle is u (at t = 0) and the acceleration a is given by alpha t^3/2. Which of the following relations is valid?

(1) v=u + alpha t^3/2

(2)v=u+ 3alpha t^3/2

(3) v=u+2/5alpha t^5/2

(4)v =u + alpha t^5/2


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Answer 1

Answer:

Option (3)

Explanation:

We know that

$a = \frac{dv}{dt}$

And,

$\\ a = \alpha {t}^{ \frac{3}{2}}$

$\rule{200}{2}$

So,

$\frac{dv}{dt} = \alpha {t}^{ \frac{3}{2} }$

Now,

Integrating both the sides :-

$\int{\dfrac{dv}{dt}} = \int{ \alpha {t}^{ \frac{3}{2} }}\\ \\ \\\int{\dfrac{dv}{dt}} = \alpha \int{ {t}^{ \frac{3}{2} }} \\ \\ \\\int{\dfrac{dv}{dt}} = \alpha ( \frac{ {t}^{( \frac{3}{2} + 1)} }{ \frac{3}{2} + 1}) + c \\ \\ \\ \\v = c + \frac{2}{5} \alpha {t}^{ \frac{5}{2} }$

$\rule{200}{2}$

Here in this case the constant will be equal to the initial velocity of the object at time=0

Hence,

We get :-

$v = u + \frac{2}{5} \alpha {t}^{ \frac{5}{2} }$

$\rule{200}{2}$

Answer 2

The attachment below explains the answer