Question

A car traveling at 43 ft/sec decelerates at a constant 7 feet per second per second. How many feet does the car travel before coming to a complete stop

Answer 1

Answer: 132.07 feet approximately

Work Shown:

• vi = initial velocity = 43 ft per sec
• vf = final velocity = 0 ft per sec, since we want the car to stop
• a = acceleration = -7 ft/s per sec, negative acceleration means we slow down
• d = unknown stopping distance in feet

Solve for d.

(vf)^2 = (vi)^2 + 2*a*d

(0)^2 = (43)^2 + 2*(-7)*d

0 = 1849 + -14*d

-1849 = -14*d

d = (-1849)/(-14)

d = 132.071428571429 approximately

d = 132.07 feet approximately

Round this however your teacher instructs.

For more info, search out "kinematics equations".