# Kinematics car braking scenario

1. Oct 30, 2016

### late347

1. The problem statement, all variables and given/known data
car is braking evenly. The brakes are in locking essentially (not ABS brakes) Car is sliding under sliding friction. Car comes to a fullstop at the distance of 150m
starting velocity was 100km/h (27,77m/s)

calculate
a) braking time
b)cars deceleration
c) friction coefficient between tyre and road

known variables
s=150m
mass=1500kg
$V_0=27,77 m/s$
$V_1= 0 m/s$
$t_0= 0s$
$t_1= ???$
$a= ???$
$F= ???$

2. Relevant equations
$V_{avg} * t = s$
$V_1= V_0 +at$
$F=ma$
$F=F_μ$
$a=μ*g$
$F_μ= μ* F_N$
$|F_N|=|G|$

$G=mg$
3. The attempt at a solution

This was an old exam problem the most difficult of them I would say.

I did not do it very well, under the time pressure. But I attempted to do it at home afterwards when I hopefull was able to do better quality creative thinking.

It is known that car decelerates at even acceleration.
therefore (it took me some considerable time to realize how to calculate the braking time, this was tough cookie for me)

$s= V_{avg} * t_1$
$t_1=10.8030s$

thus
$V_1=V_0+at$
$a= -2,5705 m/s^2$

thus
after realizing that F_net upon the car in the horizontal axis should equal the force of braking. Gravity and support force cancel out.

$F_μ=ma$

apparently according to my notes from lecture, it seems that
thusly
$a= μ*(g)$
$-2,5705m/s^2= -9.81m/s^2 *μ$
$μ=0,2620$

2. Oct 30, 2016

### PeroK

3. Oct 30, 2016

### Staff: Mentor

Yes, that all looks good.

4. Oct 30, 2016

### late347

the point about the calculate friction coefficient was more difficult.

I think it was more about the absolute value signs used in my teacher's notes that he gave us.

$a=|\vec{a}|= \frac{|∑\vec{F}|}{m}=\frac{|\vec{F_{μ}}|}{m}= \frac{μ*|\vec{F_{N}}|}{m}=\frac{μ*|\vec{G}|}{m}=\frac{μmg}{m}=μg$

Is this really true and how is it true, especially given the things about absolute values in the equation.
$\frac{|\vec{F_{μ}}|}{m}= \frac{μ*|\vec{F_{N}}|}{m}$

5. Oct 30, 2016

### Staff: Mentor

Using the magnitudes of the vectors simplifies things. We know, for example that the normal to the surface where the motion is taking place must be at right angles to the motion. We also know that (kinetic) friction always acts opposite to the direction of motion, and has the same magnitude regardless of its direction. It's much simpler to deal with friction as a magnitude and assign the direction to the friction force when we know the direction of motion.

Does that address your issue? I wasn't sure exactly what was giving you problems.

6. Oct 30, 2016

### late347

did an alternate strategy exist for solving the braking time?

(some other method than that which I used with average velocity times the time = distance)

7. Oct 30, 2016

### Staff: Mentor

Sure. You could have found the acceleration first from the initial velocity and distance with an appropriate choice of equation from the SUVAT menu. Then use the one relating distance, acceleration, and time to extract the time.

8. Oct 31, 2016

### late347

seeing as how I already did the problem in one way successfully, I can ask for clarification.

care to elaborate on which equation you would use to get the acceleration calculated?

Because you don't know acceleration and neither the ending time.

I think the problem certaniily seems to guide you towards first getting the time from
$s= V_{avg} * t$
$V_{avg}=\frac{V_1+V_0}{2}$

then doing the other calculations. But I suppose in the end the two other(?) SUVATs can be derived from only a couple of those equations. It seems the more fundamental ones were the
$v=u+at$
$s= \frac{v+u}{2}*t$ for distance in even acc

9. Oct 31, 2016

### Staff: Mentor

I'd use $v_f^2 - v_i^2 = 2 a s$.