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I'm trying to find the maximum displacement of a object given the trajectory. The trajectory is given by the equation: $$p(t)=-11.79t^2+25.9t+4.35$$ Looking around online, the closest article I've seen says that the maximum displacement is when $v(t)=0$, and velocity is the derivative of the trajectory (position as a function of time). what I have done is: $$v(t)=\frac d{dt}[-11.79t^2+25.9t+4.35]$$ $$v(t)=-23.58t+25.9$$ $$0=-23.58t+25.9$$ $$t=\frac{25.9}{23.58}=1.098388465$$ Then plugging this into the original equation gives the max displacement as $$displacement=47.02239186$$

Is this even remotely correct? If not, how would one go about finding the max displacement given a trajectory? Thank you in advance.

Tom-Tom
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Alex Anderson
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1 Answers1

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You have done exactly what was expected, but when you plugged your $t$ into the displacement equation you made an error. Just plugging in $t=1$ we have $p(1)=18.46$

Ross Millikan
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  • You're correct. When plugging it into the calculator I forgot the negative in front of the first term. Thank you for your help. – Alex Anderson Sep 22 '15 at 02:53