Question: Starting from rest, acceleration of a particle is $a = 2(t - 1).$ The velocity of the particle at ...

Question

Starting from rest, acceleration of a particle is $a = 2(t - 1).$ The velocity of the particle at $t = 5s$ is

Answer 1

Solution

$\because a = \frac{dv}{dt} = 2(t - 1) \Rightarrow dv = 2(t - 1)\mspace{6mu} dt$

$\Rightarrow \nu = \int_{0}^{5}2\mspace{6mu}(t - 1)dt = 2\left\lbrack \frac{t^{2}}{2} - t \right\rbrack_{0}^{5} = 2\left\lbrack \frac{25}{2} - 5 \right\rbrack$ = 15 m/s

Question: Starting from rest, acceleration of a particle is \(a = 2(t - 1).\) The velocity of the particle at ...

Solution