A car traveling at 46 mph decelerates at 21 feet per second per second.
a). How long does it take for the car to come to a complete stop?
b). What distance is required to bring the car to a complete stop? An
convert 46 mph to ft/s. If that value is c, then
v = c-21t
solve for t when v=0
the distance s is
s = ct - 10.5 t^2
-51.6
5t
To find the answers to these questions, we need to use the equations of motion. Let's break down the problem step by step.
a) How long does it take for the car to come to a complete stop?
To solve for the time it takes for the car to stop, we need to find the time when its velocity reaches 0. We can use the equation:
v = u + at
where
v = final velocity (0 in this case)
u = initial velocity (46 mph, which needs to be converted to feet per second)
a = acceleration (-21 feet per second squared)
t = time (what we're solving for)
First, let's convert the initial velocity from mph to feet per second.
1 mph = 1.47 feet per second
So, the initial velocity u is 46 mph * 1.47 feet per second = 67.62 feet per second.
Plugging the values into the equation, we have:
0 = 67.62 + (-21)t
Simplifying,
-67.62 = -21t
Now, divide both sides of the equation by -21:
t = 67.62 / 21
t ≈ 3.22 seconds
Therefore, it takes approximately 3.22 seconds for the car to come to a complete stop.
b) What distance is required to bring the car to a complete stop?
To find the distance required, we can use another equation of motion:
s = ut + 0.5at^2
where
s = distance (what we're solving for)
u = initial velocity (67.62 feet per second)
a = acceleration (-21 feet per second squared)
t = time (3.22 seconds)
Plugging the values into the equation, we have:
s = 67.62 * 3.22 + 0.5 * (-21) * (3.22)^2
Simplifying,
s ≈ 68.65 - 34.28
s ≈ 34.37 feet
Therefore, approximately 34.37 feet of distance is required to bring the car to a complete stop.