If a car moving at 76.0 mph takes 500 ft to stop with uniform acceleration after its brakes are applied, how far will it take to stop under the same conditions if its initial velocity is 38.0 mph?.
To solve this problem, we can use the equations of motion under uniform acceleration. The first step is to convert the velocities from miles per hour (mph) to feet per second (ft/s) since the distances and time intervals are given in feet.
1 mph is equivalent to 1.47 ft/s, so we can convert 76.0 mph to ft/s by multiplying it by 1.47:
76.0 mph * 1.47 ft/s = 111.72 ft/s
Similarly, 38.0 mph is equivalent to 55.86 ft/s:
38.0 mph * 1.47 ft/s = 55.86 ft/s
Now we have the initial velocity, v₀ = 55.86 ft/s, and we need to find the stopping distance, s. We are given that the car takes 500 ft to stop with uniform acceleration when its initial velocity is 111.72 ft/s.
The equation that relates the initial velocity (v₀), final velocity (v), acceleration (a), and distance (s) is:
v² = v₀² + 2as
We can rearrange this equation to solve for s:
s = (v² - v₀²) / (2a)
Since the car is slowing down, the final velocity (v) will be 0. Plugging in the values we have:
s = (0² - 55.86²) / (2a)
To find the acceleration (a), we can use the equation that relates acceleration, final velocity, and time:
v = v₀ + at
Since the final velocity (v) is 0, we can solve for a:
0 = 55.86 + a * t
a * t = -55.86
a = -55.86 / t
Using the first scenario where v₀ = 111.72 ft/s and s = 500 ft, we can solve for t:
500 = (0² - 111.72²) / (2 * a₁)
a₁ = (-111.72²) / (2 * 500)
Now that we know a₁, we can find t:
t = -55.86 / a₁
Solving this equation will give us the time it takes for the car to stop under the first scenario.
Finally, with the acceleration (a₁) and time (t) calculated above, we can find the stopping distance (s₂) when the initial velocity (v₀) is 55.86 ft/s using the same equation as before:
s₂ = (0² - 55.86²) / (2 * a₁)
By substituting the values into the equations and solving them, we can find the distance it will take to stop under the same conditions if the initial velocity is 38.0 mph.