Question

An engineer is setting a speed limit on a road. She wants a car to be able to stop in 147 ft and knows that the maximum rate at which a car can safely brake is 20.0 ft/s². What initial speed (in given units) will allow the car to come to a stop at the stop sign when braking with this acceleration?

Answer 1

The engineer needs to set a speed limit that allows the car to stop in 147 ft. The maximum deceleration rate is 20.0 ft/s². The initial speed required for the car to decelerate and come to a stop at the stop sign is approximately 76.63 ft/s.

Step-by-step explanation:

To calculate the initial speed at which the car will be able to stop at the stop sign when braking with a maximum acceleration of 20.0 ft/s², we can use the kinematic equation:

v² = u² + 2as

where v is the final velocity (0 ft/s), u is the initial velocity (unknown), a is the acceleration (-20.0 ft/s²), and s is the distance traveled (-147 ft).

Plugging in the known values, we get:

0 = u² + 2(-20.0)(-147)

Solving for u², we have:

u² = 2(20.0)(-147)

u² = -5880

Taking the square root of both sides, we find:

u = ±76.63 ft/s

Since we are only interested in the positive value (as the car cannot have a negative initial speed), the initial speed required for the car to come to a stop at the stop sign is approximately 76.63 ft/s.

Answer 2

The initial speed that will allow the car to come to a stop at the stop sign when braking with an acceleration of -20.0 ft/s² is approximately 76.65 ft/s.

Step-by-step explanation:

To calculate the initial speed, we can use the equation:

Vf^2 = Vi^2 + 2ad

Where Vf is the final velocity (0 m/s since the car comes to a stop), Vi is the initial velocity (what we want to find), a is the acceleration (-20.0 ft/s²), and d is the distance (147 ft). Rearranging the equation, we get:

Vi^2 = Vf^2 - 2ad

Plugging in the values, we find:

Vi^2 = 0^2 - 2(-20.0 ft/s²)(147 ft)

Vi^2 = 5880 ft²/s²

Taking the square root of both sides, we get:

Vi = sqrt(5880) ft/s

Using a calculator, we find that Vi ≈ 76.65 ft/s. Therefore, the initial speed that will allow the car to come to a stop at the stop sign when braking with this acceleration is approximately 76.65 ft/s.