When the driver applies the brakes of a light truck traveling at [itex]40\,\frac{km}{hr}[/itex], it skids 3 m before stopping. How far will the truck skid if it is traveling [itex]80\,\frac{km}{hr}[/itex] when the brakes are applied?(adsbygoogle = window.adsbygoogle || []).push({});

Here is what I have done, we are supposed to use the work energy formula, but I can't figure out how to relate the small amount of information. There is no mass specified, no kinetic friction coeeficient specified? Please help!

[tex]V_{i_1}\,=\,40\,\frac{km}{hr}\,=\,11.11\,\frac{m}{s}[/tex]

[tex]V_{f_1}\,=\,0[/tex]

I figured [itex]t_1[/itex] using kinematics:

[tex]v_f\,=\,v_i\,+a\,t[/tex]

[tex]0\,=\,11.1\,+\,a\,t[/tex]

[tex]s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a\,t^2[/tex]

[tex]6\,=\,0\,+\,22.2\,+\,a\,t^2[/tex]

[tex]t_1\,=\,0.54\,s[/tex]

[tex]a\,=\,-20.6\,\frac{m}{s^2}[/tex]

Then I use these numbers in another kinematic equation for the 80 km/hr instance:

[tex]V_f\,=\,V_0\,+\,2\,a\,(s\,-\,s_0)[/tex]

[tex]0\,=\,22.2\,\frac{m}{s}\,+2\,\left(-20.6\,\frac{m}{s^2}\right)\,(s\,-\,0)[/tex]

[tex]s\,=\,0.538\,m[/tex]

The answer is actually 12m though.

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# Homework Help: Driver going 40km/hr slams on the brakes & goes 3m before stopping - how far for 80?

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