Question

A body starts from rest, with uniform acceleration a, the acceleration of the body as function of
time t' is given by the equation a = pt, where p is the constant, then the displacement of the
particle in the time intervalt = 0 to t= t, will be
1. pt
2. -pt
3. pt?
4 pt
-
+3​

Answer 1

Answer:

$p {t}^{3} \div 6$

Explanation:

Refer to the material.

Answer 2

Answer:

The displacement of the  particle in the time interval t = 0 to t= t is  $\dfrac{pt^3}{6}$ .

Explanation:

Given :

Acceleration of body is , a = pt ( Here , p is a constant ) .

We need to find the displacement of the  particle in the time interval t = 0 to t= t .

We know ,

$v=\int\limits^t_0 {a} \, dt\\\\v=\int\limits^t_0 {pt} \, dt\\\\v=p|\dfrac{t^2}{2}|_0^t\\\\v=\dfrac{pt^2}{2}$( Here , v is velocity )

Also , displacement d is :

$d=\int\limits^t_0 {\dfrac{pt^2}{2}} \, dt\\ \\d=p|\dfrac{t^3}{6}|_0^t\\\\d=\dfrac{pt^3}{6}$

Therefore , the displacement of the  particle in the time interval t = 0 to t= t is  

$\dfrac{pt^3}{6}$ .