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Can someone help me with some physics questions and explain to me how you did them?
53. A skateboarder rolls down a hill with an acceleration of 4.0 m/s*2. He is on a hill for 4.8 s. What is his change in speed?
54. A train is moving at 15.0 m/s. After 12.0 s, it is moving at 35 m/s. What is the train's rate of acceleration?
55. A skier accelerates with an average acceleration of 6.0 m/s*2. Near the top of the hill her speed was 15.0 m/s. Calculate the time it took her to change her speed to 16.5 m/s near the bottom of the hill.
2 Answers
- JimLv 78 years ago
53. Change in speed = acceleration multiplied by the time
54. Change in speed divided by time = acceleration*
55. Change in speed divided by acceleration = time
Comment: incorrect wording in 54:
The train's acceleration is what should be asked for... not the RATE of acceleration.
RATE of acceleration means the time rate of acceleration or what is officially known as "jerk". This quantity is not usually calculated or used in an elementary (1st year) physics course.
- peterpanLv 78 years ago
53)
vf=vi+at
vf=vi+4*4.8=vi+19.2 m/s
Δv=vf-vi=19.2 m/s
54)
a=Δv/Δt
a=(vf-vi)/Δt
a=(35-15)/12=0.83 m/s^2
55)
vf=vi+at
then
t=(vf-vi)/a
t=(16.5-15)/6=0.25 s
