A particle moves with a constant acceleration `a=2ms^-2` along a straight line. If it moves with an initial velocity of `5ms^-1`, then obtain an expression for its instantaneous velocity.

Explanation

We know that acceleration is the time rate of change of velocity, i.e. `a=dv//dt` and differentiation is the inverse operation of integration. So by integrating acceleration, we can obtain the expression of velocity.
So, `v=intadt=2intdt=2t+c`
where c is the constant of integration and its value can be obtained from the initial conditions.
At `t=0`, `v=5ms^-1`. Putting these in Eq. (i), we have
`5=2xx0+c`
`impliesc=5ms^-1`
Therefore, `v=2t+5` is the required expression for the instantaneous velocity.