A car is traveling at 70 mph skids 240 ft after it's breaks are suddenly applied. Assuming that the cars breaking system provides a constant deceleration, find that deceleration. How long did it take the car to stop?
You want me to do this with calculus with these terrible units evidently so I will.
70 mph (1h/3600s)(5280ft/m) = 103 ft/s
d^2x/dt^2 = a
dx/dt = a t + Vi where Vi is initial speed
x = (1/2) a t^2 + Vi t + Xo wher Xo is initial x (call it 0 here
so
dx/dt at end = 0
0 = a t + 70 so a = -103/t
then
240 ft = (1/2) at^2 + 103 t
240 = (1/2)(-103/t) t^2 + 103 t
240 = -51.5 t + 103 t = 51.5 t
t = 4.66 seconds to stop
a = -103/4.66 = -22.1 ft/s^2
240 =
To find the deceleration of the car, we can use the following formula:
v^2 = u^2 + 2as
where:
v = final velocity (0 mph, since the car stops)
u = initial velocity (70 mph)
a = deceleration (unknown)
s = distance traveled (240 ft)
First, we need to convert the units from mph and ft to a more standard unit like meters per second (m/s) and meters (m). Here are the necessary conversions:
70 mph = 31.2928 m/s (1 mph = 0.44704 m/s)
240 ft = 73.152 m (1 ft = 0.3048 m)
Now, we can plug in the values into the formula:
(0 m/s)^2 = (31.2928 m/s)^2 + 2(a)(73.152 m)
Simplifying the equation:
0 = 976.5625 m^2/s^2 + 146.304a
Next, we can isolate the deceleration (a) term by rearranging the equation:
146.304a = -976.5625 m^2/s^2
Now, solve for a by dividing both sides of the equation by 146.304:
a = -6.678 m/s^2 (rounded to three decimal places)
Therefore, the deceleration of the car is approximately -6.678 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity (deceleration in this case).
To determine the time it takes for the car to stop, we can use the following equation:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (31.2928 m/s)
a = deceleration (-6.678 m/s^2)
t = time (unknown)
Rearranging the equation to solve for t:
0 = 31.2928 m/s + (-6.678 m/s^2) * t
31.2928 m/s = 6.678 m/s^2 * t
t = 31.2928 m/s / 6.678 m/s^2
t ≈ 4.691 seconds
Therefore, it takes approximately 4.691 seconds for the car to stop.