In the Soapbox Derby, young participants build non-motorized cars with very low-
ID: 1432409 • Letter: I
Question
In the Soapbox Derby, young participants build non-motorized cars with very low-friction wheels. Cars race by rolling down a hill. The track at Akron's Derby Downs, where the national championship is held, begins with a 55ft-long section tilted 13 below horizontal.
What is the maximum possible acceleration of a car moving down this stretch of track?
If a car starts from rest and undergoes this acceleration for the full l, what is its final speed in m/s?
Express your answer to two significant figures and include the appropriate units.
In the Soapbox Derby, young participants build non-motorized cars with very low-friction wheels. Cars race by rolling down a hill. The track at Akron's Derby Downs, where the national championship is held, begins with a 55ft-long section tilted 13 below horizontal.
What is the maximum possible acceleration of a car moving down this stretch of track?
If a car starts from rest and undergoes this acceleration for the full l, what is its final speed in m/s?
Express your answer to two significant figures and include the appropriate units.
v =Explanation / Answer
First change 55 ft to meters,
So, 55ft x (12in/1ft) x (2.54cm/1in) x (1m/100cm) = 16.7m
a) On an inclined plane, acceleration = g x sin(angle) with g=9.8
a = (9.8)(sin(13))
a= 2.2m/s^2 ...............Ans.
b) First, we have given and from a:
a = 2.2m/s^2
deltaX (or l) = 16.7 m
initial velocity = 0m/s (because it starts from rest)
we know that
v(final)^2 = v(initial)^2 + 2(a)(deltaX)
because V(initial) = 0, the equation is v(final)^2 = 2(a)(deltaX)
v^2=2(2.2m/s^2)(16.7m)
v^2 = 73.48m^2/s^2
v = sqr rt (73.48m^2/s^2)
v = 8.57 m/s.............Ans.