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

VinnyCee

When the driver applies the brakes of a light truck traveling at $40\,\frac{km}{hr}$, it skids 3 m before stopping. How far will the truck skid if it is traveling $80\,\frac{km}{hr}$ 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!

$$V_{i_1}\,=\,40\,\frac{km}{hr}\,=\,11.11\,\frac{m}{s}$$

$$V_{f_1}\,=\,0$$

I figured $t_1$ using kinematics:

$$v_f\,=\,v_i\,+a\,t$$

$$0\,=\,11.1\,+\,a\,t$$

$$s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a\,t^2$$

$$6\,=\,0\,+\,22.2\,+\,a\,t^2$$

$$t_1\,=\,0.54\,s$$

$$a\,=\,-20.6\,\frac{m}{s^2}$$

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

$$V_f\,=\,V_0\,+\,2\,a\,(s\,-\,s_0)$$

$$0\,=\,22.2\,\frac{m}{s}\,+2\,\left(-20.6\,\frac{m}{s^2}\right)\,(s\,-\,0)$$

$$s\,=\,0.538\,m$$

The answer is actually 12m though.

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d_leet

VinnyCee said:
When the driver applies the brakes of a light truck traveling at $40\,\frac{km}{hr}$, it skids 3 m before stopping. How far will the truck skid if it is traveling $80\,\frac{km}{hr}$ 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!

$$V_{i_1}\,=\,40\,\frac{km}{hr}\,=\,11.11\,\frac{m}{s}$$

$$V_{f_1}\,=\,0$$

I figured $t_1$ using kinematics:

$$v_f\,=\,v_i\,+a\,t$$

$$0\,=\,11.1\,+\,a\,t$$

$$s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a\,t^2$$

$$6\,=\,0\,+\,22.2,+\,a\,t^2$$
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

VinnyCee

$$v_f^2\,-\,v_0^2\,=\,2\,a\,s$$

$$(0)^2\,-\,\left(22.2\,\frac{m}{s}\right)^2\,=\,2\,\left(-20.6\,\frac{m}{s^2}\right)\,s$$

$$-492.8\,\frac{m^2}{s^2}\,=\,-41.2\,\frac{m}{s^2}$$

$$s\,=\,\frac{-492.8\,\frac{m^2}{s^2}}{-41.2\,\frac{m}{s^2}}\,=\,11.96m$$

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. - $\frac{1}{2}\,m\,v_0^2\,+\,\sum\,U\,=\,\frac{1}{2}\,m\,v_f^2$) somehow.

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d_leet

VinnyCee said:
$$v_f^2\,-\,v_0^2\,=\,2\,a\,s$$

$$(0)^2\,-\,\left(22.2\,\frac{m}{s}\right)^2\,=\,2\,\left(-20.6\,\frac{m}{s^2}\right)\,s$$

$$-492.8\,\frac{m^2}{s^2}\,=\,-41.2\,\frac{m}{s^2}$$

$$s\,=\,\frac{-492.8\,\frac{m^2}{s^2}}{-41.2\,\frac{m}{s^2}}\,=\,11.96m$$

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

arunbg

Note that you don't need to actually find the deceleration to get to the answer. This can be done as follows,
$$v^2 = 2aS=>v^2\varpropto S$$
since acceleration is constant in both cases , we get
$$\frac{{v_1}^2}{{v_2}^2}=\frac{S_1}{S_2}$$
Now you can solve by inserting given values.

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VinnyCee

$$\frac{40\frac{km}{hr}}{80\frac{km}{hr}}\,=\,\frac{3\,m}{X}$$

$$X\,=\,6\,m$$

But it's really 12!

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Hootenanny

Staff Emeritus
Gold Member
VinnyCee said:
$${\color{red}\frac{40\frac{km}{hr}}{80\frac{km}{hr}}}\,=\,\frac{3\,m}{X}$$

$$X\,=\,6\,m$$

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

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