A car is traveling at 8.9 m/s (20mph) in a school zone. A child darts in front of the car. Ignoring reaction time, the car can accelerate (brake) at -4.57 m/s.
-How long will it take for the car to stop?
-How far will the car travel in that time?
t = 8.9 / 4.57
stopping distance = (8.9 / 2) * t
To calculate the time it takes for the car to stop, we need to use the formula:
\( time = \frac{final\ velocity - initial\ velocity}{acceleration} \)
where the final velocity is 0 m/s since the car will stop, the initial velocity is 8.9 m/s, and the acceleration is -4.57 m/s² (negative because it represents deceleration or braking).
Substituting the values into the formula:
\( time = \frac{0\ m/s - 8.9\ m/s}{-4.57\ m/s²} \)
Simplifying the equation further:
\( time = \frac{-8.9\ m/s}{-4.57\ m/s²} \)
Dividing -8.9 by -4.57:
\( time \approx 1.947\ seconds \)
Therefore, it will take approximately 1.947 seconds for the car to stop.
To calculate the distance the car will travel during this time, we can use the formula:
\( distance = \frac{1}{2} \times acceleration \times time^2 + initial\ velocity \times time \)
Substituting the values into the formula:
\( distance = \frac{1}{2} \times -4.57\ m/s² \times (1.947\ s)^2 + 8.9\ m/s \times 1.947\ s \)
Simplifying the equation:
\( distance = \frac{1}{2} \times -4.57\ m/s² \times 3.790\ s^2 + 8.9\ m/s \times 1.947\ s \)
Calculating the squares and multiplying:
\( distance = -4.57\ m/s² \times 7.242\ s² + 8.9\ m/s \times 1.947\ s \)
Finally, combining the terms:
\( distance = -33.08\ m + 17.32\ m \)
\( distance = -15.76\ m \)
Therefore, the car will travel approximately 15.76 meters (rounded to two decimal places) during this time. Note that the negative sign indicates that the car traveled in the opposite direction of its initial velocity.