Question

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 40 ft/s2. What is the distance (in ft) traveled before the car comes to a stop?

Answer 1

To find the distance traveled before the car comes to a stop, we can use the following kinematic equation:

\[
v^2 = u^2 + 2a s
\]

where:
- \( v \) is the final velocity (0 ft/s, since the car stops),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (deceleration, so it will be negative), and
- \( s \) is the distance traveled.

First, we need to convert the initial speed from miles per hour to feet per second. We know that:

1 mile = 5280 feet,
1 hour = 3600 seconds.

Thus, to convert 50 mi/h:

\[
u = 50 \text{ mi/h} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 50 \times \frac{5280}{3600}
\]
\[
u = 50 \times 1.4667 \approx 73.33 \text{ ft/s}
\]

Now substituting the values into the kinematic equation. Since the deceleration is 40 ft/s\(^2\), we set \( a = -40 \text{ ft/s}^2 \):

\[
0 = (73.33)^2 + 2(-40) s
\]
\[
0 = 5377.6889 - 80s
\]
\[
80s = 5377.6889
\]
\[
s = \frac{5377.6889}{80} \approx 67.22 \text{ ft}
\]

The distance traveled before the car comes to a stop is approximately **67.22 feet**.