A driver in a car traveling at a speed of
59 mi/h sees a deer 104 m away on the road.
Calculate the minimum constant acceler-
ation that is necessary for the car to stop
without hitting the deer (assuming that the
deer does not move in the meantime).
Answer in units of m/s2.
Use v^2=u^2+2as
so a=(v^2-u^2)/2s
final v=0 m s^-1
initial u you will need to convert from 59 mi/h to m s^-1
s=104 m
hence calculate a
To solve this problem, we can use the kinematic equation that relates speed, distance, and acceleration:
v^2 = u^2 + 2ad
Where:
v = final velocity (0 m/s, since the car needs to stop)
u = initial velocity (59 mi/h converted to m/s)
a = acceleration
d = displacement (104 m)
First, we need to convert the initial velocity from miles per hour (mi/h) to meters per second (m/s):
1 mile = 1609.34 meters
1 hour = 3600 seconds
So, 59 mi/h = (59 * 1609.34 m) / (3600 s) ≈ 26.3867 m/s
Now, we plug the values into the equation and solve for acceleration (a):
0^2 = (26.3867 m/s)^2 + 2a(104 m)
This simplifies to:
0 = 694.642 m^2/s^2 + 208a
Next, we isolate the acceleration term:
-208a = 694.642
a = 694.642 / -208
a ≈ -3.34 m/s^2
The minimum constant acceleration necessary for the car to stop without hitting the deer is approximately -3.34 m/s^2. The negative sign indicates that the acceleration should be in the opposite direction of the car's initial velocity.