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.
4 answers