## Related questions with answers

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

VerifiedFirst 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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