A car traveling at 50 ft/sec decelerates at a constant 8 feet per second per second. How many feet does the car travel before coming to a complete stop?

complete-stop

#1

A car traveling at 50 ft/sec decelerates at a constant 8 feet per second per second. How many feet does the car travel before coming to a complete stop?

Answer:

Using the first equation of motion
v=u+a*t
u=50
a=-8
v=0 (Complete stop)
therefore,
t=( v - u ) / a
t= (0 - (-50) ) / (-8)
t=6.25 second

Now,
Using second equation of motion,

s=ut +1/2a*t^2

s=50*(6.25) + 1/2*(-8)*((6.25^2)
s=312.5 -156.25

answer= 156.25 feet