Question

A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36 m/s2 and q = -1.20 m/s .
Determine the mouse's average speed between t=1.0s and t=4.0s

Answer 1

The mouse's average speed between t=1.0s and t=4.0s is $0.6m/s$

The mouse's position as a function of time is given by ,

                               $x(t)=pt^{2}+qt$

Substitute p = 0.36 and q = -1.20 in above relation.

             $x(t)=0.36t^{2}-1.2t$

            $x(1)=0.36-1.2=-0.84m\\\\x(4)=5.76-4.8=0.96m$

the mouse's average speed between t=1.0s and t=4.0s is,

                   $v=\frac{x(4)-x(1)}{4-1} =\frac{0.96+0.84}{3}=0.6m/s$

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