Light Truck Skidding Distance at Different Speeds

[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.

 

[d_leet]

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

v_f² - 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.

 

[d_leet]

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.

Your welcome, I'm glad I could 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.
Do you follow ?

 

[VinnyCee]

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

X\,=\,6\,m

But it's really 12!

 

[Hootenanny]

{\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.