A skateboarder at a skate park rides along the path shown in (Figure 1) .
a=2.7 m and b=1.0m
If the speed of the skateboarder at point A is v=1.6m/s, what is her speed at point B? Assume that friction is negligible.
3 answers
6.71m/s
Solve the downhill speed first:
Vf = sqrt Vi^2+2*mh => Vf=sqrt 1.4m/s^2+2*(9.81)*(2.7m)=7.42m/s
Next solve uphill speed. Use Vf from downhill speed as Vi for calculating final speed going uphill.
Use the same formula as before but replace height (h) with the delta in height. We're calculating from the ground level up 1 meter so the initial height =0m, final height =1m so delta y = -1m. Makes sense this is a negative number since now the skateboarder is going uphill and losing speed.
Vf= sqrt 7.42m/s^2+2*(9.81)*(-1)= 5.95m/s
Vf = sqrt Vi^2+2*mh => Vf=sqrt 1.4m/s^2+2*(9.81)*(2.7m)=7.42m/s
Next solve uphill speed. Use Vf from downhill speed as Vi for calculating final speed going uphill.
Use the same formula as before but replace height (h) with the delta in height. We're calculating from the ground level up 1 meter so the initial height =0m, final height =1m so delta y = -1m. Makes sense this is a negative number since now the skateboarder is going uphill and losing speed.
Vf= sqrt 7.42m/s^2+2*(9.81)*(-1)= 5.95m/s
Mistake in that answer.
First equation should have read: Vf = sqrt Vi^2+2*gh. Sorry.
First equation should have read: Vf = sqrt Vi^2+2*gh. Sorry.
1 answer