Question

Haley is driving down a straight highway at 65 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?

Express your answer in meters per second squared.

Answer 1

Haley will need to decelerate at a constant rate of approximately -0.4927 m/s² to reduce her speed from 65 mph to 55 mph over the distance of 0.50 miles.

Step-by-step explanation:

To determine what constant acceleration will bring Haley to the lower speed limit of 55 mph in the distance of 0.50 miles, we can use the kinematic equation:

v2 = u2 + 2as

Where:

v is the final velocity (55 mph, converted to meters per second),

u is the initial velocity (65 mph, converted to meters per second),

a is the acceleration (in meters per second squared),

s is the distance over which the acceleration takes place (0.50 miles, converted to meters).

First, we need to convert the speeds from mph to m/s:

• 65 mph = 29.0576 m/s (approximately)
• 55 mph = 24.5872 m/s (approximately)

And the distance from miles to meters:

• 0.50 miles = 804.672 meters

Now, we rearrange the kinematic equation to solve for acceleration 'a':

a = (v2 - u2) / (2s)

Plugging the values into the equation gives us:

a = (24.58722 - 29.05762) / (2 * 804.672)

a = (-4.47042) / 1609.344 = -0.4927 m/s2 (approximately)

Haley will need to decelerate at a constant rate of approximately -0.4927 m/s2 to reduce her speed from 65 mph to 55 mph over a half-mile distance.