The Notes: A car travelling at 22.4m/s skids to stop in 2.50s. Determine the skidding distance of the car and its uniform acceleration.

Monday, April 5, 2010

A car travelling at 22.4m/s skids to stop in 2.50s. Determine the skidding distance of the car and its uniform acceleration.

Hello!

The acceleration (actually, deceleration) of a car is assumed to be constant. This is more or less correct because of Newton's Second law, $\displaystyle{a}=\frac{{F}}{{m}},$ and the breaking force $\displaystyle{F}$ is more or less constant.

Denote the magnitude of acceleration as $\displaystyle{a}$ (a positive number). Also denote the initial speed as $\displaystyle{V}_{{0}}$ and the time of a stop as $\displaystyle{t}_{{1}}.$

The speed $\displaystyle{V}$ changes uniformly with a time, the formula is obviously $\displaystyle{V}{\left({t}\right)}={V}_{{0}}-{a}{t}.$ The time $\displaystyle{t}_{{1}}$ when a car is stopped is that moment when $\displaystyle{V}{\left({t}_{{1}}\right)}={0},$ i.e. $\displaystyle{V}_{{0}}={a}{t}_{{1}}$ and $\displaystyle{a}=\frac{{V}_{{0}}}{{t}_{{1}}}=\frac{{22.4}}{{2.50}}={8.96}{\left(\frac{{m}}{{{s}}^{{2}}}\right)}.$

The distance $\displaystyle{D}$ after the skidding begins is $\displaystyle{D}{\left({t}\right)}={V}_{{0}}{t}-\frac{{{a}{{t}}^{{2}}}}{{2}},$ so the total skidding distance is $\displaystyle{D}{\left({t}_{{1}}\right)}={V}_{{0}}{t}_{{1}}-\frac{{\frac{{V}_{{0}}}{{t}_{{1}}}\cdot{{t}_{{1}}^{{2}}}}}{{2}}=\frac{{{V}_{{0}}{t}_{{1}}}}{{2}}={28}{\left({m}\right)}.$