A driver in a car traveling at a speed of 55.4 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).
I simply do not know how to set this up. Initial velocity is given, as well as distance. But how am I supposed to find anything else? I tried using final velocity as 0 and that didn't work.
First unify the units.
55.4 mi/hr = 24.8 m/s
you know that the distance is given by
s = 24.8t - 1/2 at^2
So, to stop in exactly 116 meters, we need
24.8t - 1/2 at^2 = 116
Now, we know that during that time, the speed reduces from 24.8 m/s to zero, so
a = 24.8/t m/s^2
t = 24.8/a
Thus we need
24.8(24.8/a) - 1/2 a (24.8/a)^2 = 116
a = 2.65 m/s^2
check:
t = 24.8/2.65 = 9.36 s
24.8(9.36) - 1/2 (2.65)(9.36^2) = 116.04
close enough for jazz
To solve this problem, you can use one of the kinematic equations of motion, which relates the initial velocity (v₀), final velocity (v), acceleration (a), and distance (d). The equation you can use is:
v² = v₀² + 2ad
In this case, the car needs to stop, so the final velocity (v) is 0. The initial velocity (v₀) is given as 55.4 mi/h. The distance (d) is given as 116 m. And you need to find the minimum constant acceleration (a).
Substituting the given values into the equation, you get:
0² = (55.4 mi/h)² + 2a(116 m)
Since the units for velocity should be consistent, convert the initial velocity from mi/h to m/s:
v₀ = 55.4 mi/h * (1609 m/1 mi) * (1 h/3600 s) ≈ 24.7556 m/s
Now solve for the acceleration (a):
0 = (24.7556 m/s)² + 2a(116 m)
Simplify the equation:
0 = 612.7532 m²/s² + 232a m
Rearrange the equation to solve for the acceleration (a):
232a m = -612.7532 m²/s²
Divide both sides of the equation by 232 m:
a = -2.64445 m/s²
The negative sign indicates that the car needs a negative acceleration (or deceleration) to stop in time.
Therefore, the minimum constant acceleration necessary for the car to stop without hitting the deer is approximately -2.64445 m/s².