Question

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of \(22{\rm{ ft/}}{{\rm{s}}^2}\)22 ft/s. What is the distance traveled before the car comes to a stop?

Short answer

The distance traveled before the car is about \(122.223{\rm{ ft}}\).

Step-by-step solution

Here we solve our question

We solveour question, here need to use the kinematic formula.

The kinematic formula is,

\(v_f^2 = v_i^2 + 2aD\_\_\_\_\_\_(1)\)

Here, \({v_f}\) is finale velocity which can set as 0.

\(\begin{aligned}{c}{v_i} &= {\rm{Initial velocity}}\\ &= 50{\rm{ mi/h}}\end{aligned}\)

\(\begin{aligned}{c}a &= {\rm{ Accelaration}}\\ &= 22{\rm{ ft/s}}\end{aligned}\)

And \(D\) is the distance traveled.

Find the distance

Now first here we need to change \(50{\rm{ mi/h}}\) into \({\rm{ft/s}}\) as,

\(\begin{aligned}{c}50{\rm{ mi/h}} &= \left( {50\frac{{{\rm{mi}}}}{{{\rm{1 hr}}}}} \right) \times \left( {\frac{{5280{\rm{ ft}}}}{{{\rm{1 mi}}}}} \right) \times \left( {\frac{{1{\rm{ hr}}}}{{3600{\rm{ s}}}}} \right)\\ &= 73.333{\rm{ }}\frac{{{\rm{ft}}}}{{\rm{s}}}\end{aligned}\)

Now put value \({v_i}\), \({v_f}\), \(a\) and \(D\) in equation (1).

\(\begin{aligned}{c}{0^2} &= {\left( {73.333\frac{{ft}}{s}} \right)^2} + 2\left( { - 22\frac{{ft}}{s}} \right)\left( D \right)\\0 &= 5377.7778 - 44\left( D \right)\\D &= \frac{{5377.7778}}{{44}}\\ &\approx 122.223\end{aligned}\)

The distance traveled before the car is about \(122.223{\rm{ ft}}\).