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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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First let's convert the speed to $\frac{\text{ft}}{\text{s}}$.

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

Since the deceleration is constant, let's integrate.

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

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

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

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