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Question

The speed of a car traveling in a straight line is reduced from 45 to 30 miles per hour at a distance of 264 feet. Find the distance in which the car can be brought to rest from 30 miles per hour, assuming the same constant deceleration.

Solution

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Answered 1 year ago
Answered 1 year ago
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First let's convert the speed to fts\frac{\text{ft}}{\text{s}}.

30miles=44fts45miles=66fts\begin{array}{rcl} 30 \frac{\text{mile}}{\text{s}}&=&44\frac{\text{ft}}{\text{s}}\\ 45 \frac{\text{mile}}{\text{s}}&=&66\frac{\text{ft}}{\text{s}} \end{array}

Since the deceleration is constant, let's integrate.

dv=adtv=at+C\begin{array}{rcl} \int dv&=&\int adt\\ v&=&at +C \end{array}

Substituting t=0t=0 and v=66 ftsv=66\ \frac{\text{ft}}{\text{s}} and rewriting the equation,

66=a0+CC=66v=at+66\begin{array}{rcl} 66&=&a\cdot 0+C \Longrightarrow C=66\\ v&=&\boxed{at+66} \end{array}

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