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You’re on a bike going 30mph (13.41m/s) and a squirrel jumps infront of you 60m infront. You apply the brakes to not hit the squirrel, what is the deceleration to not hit the squirrel?

User Saurab
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We will have that the deceleration can be determined as follows:

First, we recall that for linear accelerated motions the following is true:


v_f^2=v_i^2+2ad

So:


\begin{gathered} \Rightarrow(0m/s)^2=(13.41m/s)^2+2a(60m)\Rightarrow2a(60m)=-(13.41m/s)^2 \\ \Rightarrow a=(-(13.41m/s)^2)/(2(60m))\Rightarrow a=-1.4985675...m/s^2 \\ \\ \Rightarrow a\approx-1.5m/s^2 \end{gathered}

So, the deceleration must be approximately of 1.5 m/s^2.

User Jwsc
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