A 65.0kg runner\'s velocity, at the start of a long race,increases with time as:

Question

A 65.0kg runner's velocity, at the start of a long race,increases with time as: v(t)=(4.5m/s)[1-e^(-t/1.0s)] for 0.0s<t<4.0s Derive an expression for the force acting on the runner as afunction of time. What is the magnitude of the force acting on the runner att=2.0sec.? A 65.0kg runner's velocity, at the start of a long race,increases with time as: v(t)=(4.5m/s)[1-e^(-t/1.0s)] for 0.0s<t<4.0s Derive an expression for the force acting on the runner as afunction of time. What is the magnitude of the force acting on the runner att=2.0sec.?

Explanation / Answer

v(t)=(4.5m/s)[1-e^(-t/1.0s) acceleration a=dv/dt=(4.5m/s)e^(-t/1.0s) m/s2 Force F=ma=65*(4.5m/s)e^(-t/1.0s)=292.5e^(-t/1.0s) N At t=2s we have as, F=292.5e^(-2/1.0s) N Hence we get by it.

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