A car traveling at 20 miles/hour stops in a distance of 15 feet when its brakes are locked. Assuming that the frictional force doesn't depend on speed, that same car traveling at 60 miles/hour will stop in a distance closest to
a. 15 feet
b. 45 feet
c. 90 feet
d. 135 feet
My answer: c
To determine the stopping distance of a car when its brakes are locked at different speeds, we can use the principles of physics. Specifically, we can use the equation for stopping distance, which is given by:
Stopping distance = (initial velocity^2) / (2 * frictional deceleration)
In this case, the frictional force doesn't depend on speed, which means that the frictional deceleration is constant. Let's call this constant deceleration "a".
For the first scenario, the initial velocity is 20 miles/hour.
Converting this to feet/second:
20 miles/hour * (5280 feet/1 mile) * (1 hour/3600 seconds) = 29.33 feet/second
Using the equation for stopping distance:
15 feet = (29.33 feet/second)^2 / (2 * a)
Simplifying the equation:
15 = 29.33^2 / (2a)
For the second scenario, the initial velocity is 60 miles/hour.
Converting this to feet/second:
60 miles/hour * (5280 feet/1 mile) * (1 hour/3600 seconds) = 88 feet/second
Using the same equation for stopping distance:
Stopping distance = (88 feet/second)^2 / (2 * a)
Now we can compare the two stopping distances to determine which answer choice is closest.
The first stopping distance is:
15 = 29.33^2 / (2a)
Solving for "a":
a = 29.33^2 / (2 * 15) = 28.9624
Now we can calculate the stopping distance for the second scenario:
Stopping distance = (88 feet/second)^2 / (2 * 28.9624)
Stopping distance = 7744 / 57.9248
Stopping distance ≈ 133.71 feet
Comparing the stopping distance of 133.71 feet to the answer choices, we can see that the closest answer is 135 feet, which is option d. Therefore, the car traveling at 60 miles/hour will stop in a distance closest to 135 feet.