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

  • Thread starter VinnyCee
  • Start date
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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?

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.
 
Last edited:
1,073
1
VinnyCee said:
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?

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]
If the distance it travels before stopping is 3 meters, taking s0 as 0 like you've done then then s should be 3 not 6 since the total displacement is only 3 meters. Also it may have been easier to use a different equation from the start, one that doesn't take into account time since you aren't given a stopping time, such as

vf2 - v02 = 2as
 
489
0
[tex]v_f^2\,-\,v_0^2\,=\,2\,a\,s[/tex]

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

[tex]-492.8\,\frac{m^2}{s^2}\,=\,-41.2\,\frac{m}{s^2}[/tex]

[tex]s\,=\,\frac{-492.8\,\frac{m^2}{s^2}}{-41.2\,\frac{m}{s^2}}\,=\,11.96m[/tex]

That is right if you round up! Thanks for the help.

NOTE: I think we were supposed to solve it using a work-energy formula (i.e. - [itex]\frac{1}{2}\,m\,v_0^2\,+\,\sum\,U\,=\,\frac{1}{2}\,m\,v_f^2[/itex]) somehow.
 
Last edited:
1,073
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VinnyCee said:
[tex]v_f^2\,-\,v_0^2\,=\,2\,a\,s[/tex]

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

[tex]-492.8\,\frac{m^2}{s^2}\,=\,-41.2\,\frac{m}{s^2}[/tex]

[tex]s\,=\,\frac{-492.8\,\frac{m^2}{s^2}}{-41.2\,\frac{m}{s^2}}\,=\,11.96m[/tex]

That is right if you round up! Thanks for the help.
Your welcome, I'm glad I could help.
 
587
0
Note that you don't need to actually find the deceleration to get to the answer. This can be done as follows,
[tex]v^2 = 2aS=>v^2\varpropto S[/tex]
since acceleration is constant in both cases , we get
[tex]\frac{{v_1}^2}{{v_2}^2}=\frac{S_1}{S_2}[/tex]
Now you can solve by inserting given values.
Do you follow ?
 
Last edited:
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[tex]\frac{40\frac{km}{hr}}{80\frac{km}{hr}}\,=\,\frac{3\,m}{X}[/tex]

[tex]X\,=\,6\,m[/tex]

But it's really 12!
 
Last edited:

Hootenanny

Staff Emeritus
Science Advisor
Gold Member
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VinnyCee said:
[tex]{\color{red}\frac{40\frac{km}{hr}}{80\frac{km}{hr}}}\,=\,\frac{3\,m}{X}[/tex]

[tex]X\,=\,6\,m[/tex]

But it's really 12!
You forgot to square the velocities. Re calculate your value and you should obtain 12m.
 

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