A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 32 ft/s2. What is the distance covered before the car comes to a stop? (Round your answer to one decimal place.)?

Question

Julia Aguilar · Accepted Answer

#84.0ft#, rounded to one decimal place

Applicable Kinematic equation is #v^2-u^2=2as# where #v, u, a and s# are final velocity, initial velocity, deceleration and distance covered respectively. Now, #50mphxx(1760xx3)/3600=73.3333fps# Inserting given and calculated values #0^2-73.3333^2=2(-32)s# #=>s=73.3333^2/64# #s=84.0ft#, rounded to one decimal place

Parker Abbott · Answer

undefined To find the distance covered before the car comes to a stop, we can use the equation of motion:

$$d = \frac{v^2}{2a}$$

where $d$ is the distance covered, $v$ is the initial velocity, and $a$ is the deceleration.

Substituting the given values:

$$d = \frac{(50 \text{ mi/h})^2}{2(-32 \text{ ft/s}^2)}$$

$$d \approx \frac{(50 \times 5280/3600)^2}{2(-32)}$$

$$d \approx \frac{(73.33)^2}{2(-32)}$$

$$d \approx \frac{5377.69}{-64}$$

$$d \approx -84.03 \text{ ft}$$

However, distance cannot be negative, so we take the magnitude:

$$d \approx 84.03 \text{ ft}$$

Therefore, the distance covered before the car comes to a stop is approximately 84.0 feet.