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As a jet accelerates down a runway, its position as a function of time is given by x(t)=2t^2 + .5t^3, with x in meters and t in seconds. Determine the instantaneous speed of the jet at t=5s.
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How do I go about solving this? my first thought was to get the derivative of x(t), but i'm pretty lost
x=4t + 4.5t^2
V(t)=4t+1.5t^2 you can plug time t in here then you can get the answer.
thanks a lot! why is 1.5 still the coefficient? I thought you multiply it by the 3
i did but, i multiply .5 with 3, that makes 1.5
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oh oops I did 3*1.5 thanks again
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