A bicycle initially moving with a velocity 5.0 m s ^-1 accelerates for 5 s at a rate of 2 m s^-2 . What will be its final velocity?

As we know,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

Hence,

$\text {final vel. (v)} - \text {initial vel. (u)} \\[0.5em] = \text {Acceleration (a)} \times {\text {time (t)}}$

Given,

a = 2 m s^-2

t = 5 s

u = 5.0 m s ^-1

Substituting the values in the formula above, we get,

$\text {final vel. (v)} - 5 \text{m s}^{-1} = 2 \text{m s}^{-2} \times 5 s \\[0.5em] \text {v} - 5 \text{m s}^{-1} = 10 \text{m s}^{-1} \\[0.5em] \Rightarrow \text {v} = (10 + 5) \text{m s}^{-1} \\[0.5em] \Rightarrow \text {v} = 15 \text{m s}^{-1} \\[0.5em]$

Hence, final velocity = 15 m s ^-1.