question_answer

A person travelling on a scooter at 43.2 km/h applies the brakes, giving a deceleration of\[\text{6}.0\text{m}/{{\text{s}}^{\text{2}}}\]to his scooter. How far will it travel before stopping?

Answer:

\[u=43.2km/h=12m/s\] \[a=-6.0m{{s}^{-2}}.\] \[v=0\] Distance travelled by the person (S) = ? We know that \[{{v}^{2}},-{{u}^{2}}=2aS\] \[S=\frac{{{v}^{2}}-{{u}^{2}}}{2a}\] \[S=\frac{{{0}^{2}}-{{(12)}^{2}}}{2\times -6}=\frac{12\times 12}{12}=12m\]