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| *Vultures Knob>>>Bike Repair |
I have a question about basic kinematics, if anyone would have any idea on how to solve...? |
A bicyclist is finishing his repair on a flat tire when a friend rides by with a constant speed of 3.5m/s. Two seconds later the bicyclist hops on his bike and accelerates at 2.4m/s/s until he catches his friend. A) How much time does it take until he catches his friend? B) How far has he traveled in this time? C) What is his speed when he catches up? It looks like no one else is going to help you with this so I will take a stab at it. velocity=acceleration X time distance=1/2 acceleration X time^2 distance=velocity X time So: 3.5t+7=1/2 X2.4t^2 (since the two distances have to be equal when they meet, right?) 7t+14=2.4t^2 0=2.4t^2-7t-14 t=(7+-sqrt(49+4X2.4X14))/4.8 t~=4.28...seconds distance~=3.5 X 4.28 meters~=14.98 Meters speed~=2.4 X 4.28 meters/sec~=10.27meters/sec I believe these answers are approximately correct. |
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