Question

The driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown intrachead Part A

If the car's speed decreases at a constant rate from 67 mi/h to 30 mi/h in 30s, what is the magnitude of its ncceleration, assuming that continues to more Express your answer in miles per hour squared to two significant figures.
What distance does the car travel during the braking period Express your answer in feet to two significant figures ?

Answer 1

To find the magnitude of acceleration, we can use the formula:

acceleration = (final velocity - initial velocity) / time

Given:

Initial velocity (u) = 67 mi/h

Final velocity (v) = 30 mi/h

Time (t) = 30 s

Using the formula, we have:

acceleration = (30 mi/h - 67 mi/h) / 30 s

Simplifying the equation, we get:

acceleration = (-37 mi/h) / 30 s

To convert the units, we know that 1 mi/h = 1.46667 ft/s. Therefore,

acceleration = (-37 mi/h) / 30 s * (1.46667 ft/s / 1 mi/h)

Simplifying further, we get:

acceleration = -1.813 ft/s^2

So, the magnitude of acceleration is approximately 1.81 mi/h^2.

To find the distance traveled during the braking period, we can use the formula:

distance = (initial velocity + final velocity) / 2 * time

Plugging in the values, we have:

distance = (67 mi/h + 30 mi/h) / 2 * 30 s

Simplifying the equation, we get:

distance = (97 mi/h) / 2 * 30 s

To convert the units, we know that 1 mi/h = 1.46667 ft/s and 1 s = 1 ft/s. Therefore,

distance = (97 mi/h) / 2 * 30 s * (1.46667 ft/s / 1 mi/h) * (1 s / 1 ft/s)

Simplifying further, we get:

distance = 713.334 ft

So, the car travels approximately 713.33 feet during the braking period.