Question

A particle travels in a straight line, such that for a short time 2s ≤ t ≤ 6s, its motion is described by v=$\left( {\dfrac{4}{a}} \right)$ m/s, where a is in $m{s^{ - 2}}$. If v =6 $m{s^{ - 1}}$ when t = 2s, determine the particle's acceleration when t = 3s.

Answer 1

Hint: Velocity is the rate of displacement and acceleration is rate of change in velocity. According to the question the particle is moving with the certain magnitude of the velocity at time 2s such that we need to find the acceleration of the particle by using the process of integration.

Complete step by step solution:
 From the given data:
$v = \dfrac{4}{a}m{s^{ - 1}}$
Acceleration=? At time t=3s
Given velocity is
$\
  v = \dfrac{4}{a} \\
\implies a = \dfrac{4}{v} \\
\implies \dfrac{{dv}}{{dt}} = \dfrac{4}{v}.......(1) \\
\ $
Let us integrate on both the side
$\
  \int {vdv = \int {4dt} } \\
\implies \dfrac{{{v^2}}}{2} - 18 = 4t - 8 \\
\ $
$\
  \dfrac{{{v^2}}}{2} = 4t - 8 + 18 \\
\implies {v^2} = 8t + 20 \\
\implies v = \sqrt {8t + 20} \\
\ $
Equation 1 implies
$\
  a = \dfrac{{dv}}{{dt}} \\
\implies a = \dfrac{4}{{\sqrt {8t + 20} }} \\
\ $
The acceleration of the particle at time t = 3s
$\
  a = \dfrac{4}{{\sqrt {8(3) + 20} }} \\
\implies a = \dfrac{4}{{\sqrt {44} }} \\
\implies a = 0.603m{s^{ - 2}} \\
\ $

Note:
In the above question we need to be aware of how to apply the integration for an equation.
When the body is in motion then it describes the state of its motion and particles describe that it is moving in a straight line.
When the particle is moving with some acceleration which indicates that it must contain a certain velocity.