Question

Haley is driving down a straight highway at 73 mph. A construction sign warns that the speed limit will drop to 55 mph in 0.50 mi.

What constant acceleration (in m/s2) will bring Haley to this lower speed in the distance available?

Answer 1

To bring Haley to the lower speed in the available distance, a constant acceleration of -3.40 m/s^2 is required.

Step-by-step explanation:

To calculate the acceleration required to bring Haley to a lower speed in the given distance, we need to convert the speeds and distance to the same units. Let's convert the speeds from mph to m/s and the distance from miles to meters.

Haley's initial speed = 73 mph = 32.6 m/s

Lower speed (speed limit) = 55 mph = 24.6 m/s

Distance available = 0.50 mi = 804.67 m

Using the equation v^2 = u^2 + 2as, we can rearrange to solve for acceleration (a):

a = (v^2 - u^2) / (2s)

Plugging in the values:

a = (24.6^2 - 32.6^2) / (2 imes 804.67)

a = -3.40 m/s^2

Therefore, the constant acceleration required to bring Haley to the lower speed in the available distance is -3.40 m/s^2.