Answer to Question #89753 in Mechanics | Relativity for brittany

Question

The minimum distance required to stop a car traveling at 35 mi/hr is 40 ft. Assuming the same acceleration, what is the minimum distance required to stop a car traveling at 75 mi/hr?

Answer 1

Let's first find the deceleration of the car from the kinematic equation:

$v_f^2 = v_i^2 + 2as,$

here, $v_f = 0$ is the final velocity of the car, $v_i = 35 mi/hr$ is the initial velocity of the car, $a$ is the deceleration of the car and $s$ is the minimum distance traveled by the car until it stops.

Then, from this formula we can find the deceleration of the car:

$a = -\dfrac{v_i^2}{2s} = -\dfrac{(35 \dfrac{mi}{hr} \cdot \dfrac{5280 ft}{1 mi})^2}{2 \cdot 40 ft} = -4.27 \cdot 10^8 \dfrac{ft}{hr^2}.$

The sign minus indicates that the car decelerates.

Finally, we can find the minimum distance required to stop a car traveling at 75 mi/hr assuming the same rate of deceleration:

$v_f^2 = v_i^2 + 2as,$ $s = -\dfrac{v_i^2}{2a} = -\dfrac{(75 \dfrac{mi}{hr} \cdot \dfrac{5280 ft}{1 mi})^2}{2 \cdot (-4.27 \cdot 10^8 \dfrac{ft} {hr^2})} = 184 ft.$

Answer:

$s = 184 ft.$