A driver in a car traveling at a speed of 56.1 mi/h sees a deer 103 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).
(in m/s^2)
56.1 mi/h = 25.1 m/s
v = Vi - a t
0 = 25.1 - a t so t = 25.1/a
d = (1/2) a t^2
206 = a t^2
206 = a(630/a^2) = 630/a
a = 3.06 m/s^2
(negative of course)
To calculate the minimum constant acceleration needed for the car to stop without hitting the deer, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (which is 0 m/s since the car needs to stop)
u = initial velocity of the car (convert 56.1 mi/h to m/s)
a = acceleration
s = displacement (distance traveled by the car)
First, let's convert the initial velocity from miles per hour to meters per second. We know that 1 mile is equal to 1609.34 meters, and 1 hour is equal to 3600 seconds.
So, the initial velocity (u) is given by:
u = 56.1 mi/h * (1609.34 m / 1 mi) * (1 h / 3600 s) = 25.06 m/s
Next, we need to convert the distance given in meters to kilometers:
s = 103 m
Plugging the values into the kinematic equation, we get:
0 = (25.06 m/s)² + 2a(103 m)
Expanding and simplifying the equation:
0 = 627.5036 + 206a
Rearranging the equation:
206a = -627.5036
Dividing both sides by 206, we find:
a = -627.5036 / 206
So, the minimum constant acceleration required for the car to stop without hitting the deer is approximately -3.04 m/s².
Note: The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which is necessary for the car to come to a stop.