A 62.7 kg astronaut is holding a 6.5 kg tool pack while drifting at 12.3 m/s to the right. She throws the tool pack away so thatit is drifting at 2.4 m/s to the left. How fast is she moving after the throw? Assume that they are out in deep space with noother forces acting on them.

Respuesta :

Answer:

[tex]v_{new}=9.9ms^{-1}[/tex]

Explanation: Assuming that it took one second to throw the tool away, then we have to use the final velocity formula to get the new velocity:

[tex]v(t)=v_o+at[/tex]

One second would imply that the deceleration in the initial astronaut velocity is:

[tex]\begin{gathered} a=-2.4ms^{-2} \\ \end{gathered}[/tex]

Note the sign convention!

The new velocity of the astronaut after the throw would be now:

[tex]\begin{gathered} t=1s \\ \therefore\Rightarrow \\ v(1s)=12.3ms^{-1}+(-2.4ms^{-2})(1s)=(12.3-2.4)ms^{-1} \\ v_{new}=9.9ms^{-1} \end{gathered}[/tex]