If the driver of a car wishes to increase his speed from 20 mph to 50 mph while traveling a distance of 528 ft., what constant acceleration should he maintain?

Vo = 20mi/h * 5280Ft/mi * 1h/3600s. =

29.3 Ft/s. = Initial velocity.

V = (50/20) * 29.3 = 73.3 Ft/s.

V^2 = Vo^2 + 2a*d.
Solve for a, Ft/s^2.

To find the constant acceleration required for the driver to increase his speed from 20 mph to 50 mph while traveling a distance of 528 ft, we can use the kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (50 mph)
u = initial velocity (20 mph)
a = acceleration (unknown)
s = distance (528 ft)

First, we need to convert the velocities from mph to ft/s.

1 mph = 1.47 ft/s (approximately)

So, the initial velocity (u) would be 20 mph * 1.47 ft/s = 29.4 ft/s,
and the final velocity (v) would be 50 mph * 1.47 ft/s = 73.5 ft/s.

Now, let's rearrange the equation and solve for acceleration (a):

73.5^2 = 29.4^2 + 2a(528)

Simplifying the equation:

5402.25 = 864.36 + 1056a

Rearranging the equation to isolate the acceleration term:

5402.25 - 864.36 = 1056a

4537.89 = 1056a

Finally, divide both sides of the equation by 1056:

a = 4537.89 / 1056 ≈ 4.30 ft/s^2

Therefore, the driver should maintain a constant acceleration of approximately 4.30 ft/s^2 to increase his speed from 20 mph to 50 mph while traveling a distance of 528 ft.