A driver in a car traveling at a speed of 56.0 mi/h sees a deer 116 m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).
V=56mi/h * 1600m/mi * 1h/3600s=24.9 m/s
a = (Vf^2-Vo^2)/2d,
a = (0-(24.9)^2) / 232 = -2.67 m/s^2.
To calculate the minimum constant acceleration required for the car to stop in time, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
- v is the final velocity (0 in this case, as the car needs to stop)
- u is the initial velocity (56.0 mi/h)
- a is the constant acceleration we are trying to find
- s is the distance traveled (116 m)
First, let's convert the initial velocity from miles per hour to meters per second:
56.0 mi/h = 56.0 * 1609.34 m/3600 s = 25.0209 m/s (rounded to 4 decimal places).
Now, we plug the values into the equation and solve for a:
0^2 = (25.0209)^2 + 2 * a * 116
Since 0^2 = 0, we have:
0 = (25.0209)^2 + 2 * a * 116
Rearranging the equation, we get:
2 * a * 116 = - (25.0209)^2
Dividing both sides by 2 * 116, we obtain:
a = - (25.0209)^2 / (2 * 116)
Evaluating this expression, we have:
a ≈ -6.739 m/s^2 (rounded to 3 decimal places)
Since acceleration cannot be negative in this case (as we need positive acceleration to stop the car), we take the magnitude of the result:
|a| ≈ 6.739 m/s^2
Therefore, the minimum constant acceleration necessary to stop the car without hitting the deer is approximately 6.739 m/s^2.