Question

a scooter going due east at 10 metre per second turns right through an angle of 90 degree. if the speed of the scooter remains unchanged in taking turn, what is the change in velocity of the scooter?

Answer 1

$$\begin{gathered} \mathbf{Expression}\mathbf{}\mathbf{for}\mathbf{}\mathbf{change}\mathbf{}\mathbf{in}\mathbf{}\mathbf{velocity}\mathbf{:} \\ △\mathrm{v}=\sqrt{{{\mathrm{v}}_1}^2+{{\mathrm{v}}_1}^2-2{\mathrm{v}}_1{\mathrm{v}}_2\mathrm{cos\theta }} \\ \mathrm{where} \\ \mathrm{\theta }=\mathrm{Angle}\mathrm{between}\mathrm{the}\mathrm{velocities} \\ \mathrm{Hence}, \\ △\mathrm{v}=\sqrt{{10}^2+{10}^2-2\left(10\right)\left(10\right)\cos {90}^{\mathrm{o}}} \\ =10\sqrt{2} \\ \mathbf{=}\mathbf{14}\mathbf{.}\mathbf{14}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$