A driver in a car traveling at a speed of
41.7 mi/h sees a deer 88 m away on the road.
Calculate the magnitude of the acceleration
necessary for the car to stop just barely miss-
ing the deer (assuming that the deer does not
move in the meantime).
Answer in units of m/s2
Use SI units to make things easier.
Vo = 41.7 mph = 41.7*1609/3600 = 18.64 m/s
Vo = sqrt(2 a X)
X = 88 m
Solve for the reuired acceelration, a.
a = Vo^2/(2 X)= 1.97 m/s^2
To calculate the magnitude of the acceleration necessary for the car to stop just barely missing the deer, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s, as the car is coming to a stop)
u = initial velocity of the car (41.7 mi/h = 18.6 m/s)
a = acceleration
s = displacement (88 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the values into the equation:
a = (0^2 - 18.6^2) / (2 * 88)
Simplifying the equation:
a = (-345.96) / 176
a = -1.96 m/s^2
The magnitude of the acceleration necessary for the car to avoid hitting the deer is 1.96 m/s^2.