A motorbike moving with initial velocity 28 ms⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.

Initial velocity, $u = 28 \text{ ms}^{-1}$

Final velocity, $v = 0 \text{ ms}^{-1}$

Distance travelled, $s = 98 \text{ m}$

Acceleration of the motorbike can be calculated as:

$$v^2 = u^2 + 2as$$

$$\begin{aligned} 0^2 &= (28)^2 + 2 \times a \times 98 \\ 0 &= 784 + 196a \\ 196a &= -784 \\ a &= -4 \text{ ms}^{-2} \end{aligned}$$

Therefore, the acceleration of the motorbike is $-4 \text{ m s}^{-2}$. The negative sign shows that the motorbike is slowing down.

Time taken to come to a stop can be calculated using the equation,

$$\begin{aligned} v &= u + at \\ 0 &= 28 + (-4)t \\ 4t &= 28 \\ t &= 7 \text{ s} \end{aligned}$$

Therefore, the time taken by the motorbike to stop is 7 s.