In the soapbox derby, young participants build nonmotorized cars with very low friction wheels. Cars race by rolling down a hill. The track at Akron’s Derby Downs begins with a 55 ft long section tilted 13 degrees below horizontal.
A) What is the max possible acceleration of a car moving down this stretch of track?
B) If a car starts from rest and undergoes this acceleration for the full 55 feet, what is its final speed in m/s?
2 Answers

the component of gravity along an incline is g sin(theta), so for this incline the maximum possible acceleration is 9.8 sin 13 = 2.2m/s/s
this of course assumes no friction between the wheels and surface, but this is the maximum possible value
to find speed, use
vf^2=v0^2+2ad
vf=final velocity
v0=initial velocity =0
a=accel = 2.2m/s/s
d=distance =55 feet = 16.8m
vf^2=0^2+2*2.2*16.8
solve for vf (I get 8.6m/s/s)

If the boy drops the ball to unfastened fall, here could persist with preliminary velocity is 0 m/s very final velocity is 14.7 m/s Acceleration is (gravity =9.8 m/s2) distance = one million/2 * gt^2 = one million/2 * 9.8 * 4 = 19.6 m If the boy capacity throws the ball with preliminary velocity of 14.7 m/s, then here could persist with d = Vi *t + one million/2 gt^2 d= 14.7*2 + one million/2*9.8*4 = 40 9 m