# Homework, Brake Question

1. Nov 30, 2006

### XPX1

1. The problem statement, all variables and given/known data
Driving home from school one day you spot a ball rolling out into the street (Figure 5-18). You brake for 1.15 s, slowing your 1020 kg car from 16.0 m/s to 9.50 m/s.

a) What was the average force exerted on your car during braking?
Magnitude
_____ kN

(b) How far did you travel while braking?

2. Relevant equations
xf=xi+.5(vi+vf)t
vf^2=vi^2+2a(xf-xi)

3. The attempt at a solution

I have been unable to figure out this problem!

From the information gathered, I know the following
vi=16.0
vf=9.5
t=1.15

now I do not know where to go from here, ive tried using the above formulas to find the acceleration so I can plug it into the Fnet=MA equation, but still im kind of lost. How the heck do I find the average force exerted on the car while brakeing

2. Nov 30, 2006

### PhanthomJay

Use your relevant equations and solve for Xf and a, setting xi = 0. Then yes use f=ma to solve for the average force during braking.

3. Nov 30, 2006

### XPX1

im still kind of confused, I think the equations I need to use for acceleration are vf=vi+axt, but I still got the question wrong, even for part b where it asks me how far it went! Somebody please halp!

vf=vi+axt then
9.5=16+1.15a

-6.5=1.15a

a=-5.65 am I not correct?

4. Dec 1, 2006

### OlderDan

Your acceleration looks OK. How do you find the force? What distance did you get for b? Your equations are OK.

5. Dec 1, 2006

### XPX1

well b should be -5.65(1.15) which = -6.4975 but that is not correct for b.

and if I was to do the Fnet I would assume that Fnet=MA

Fnet= 1020(-5.65)

right?

6. Dec 1, 2006

### OlderDan

The force is correct once you get the units included. Multiply it out. The distance is not correct. Keeping track of units will help you know when things are not right. What you calculated is acceleration times time, which has units of velocity and is in fact the change in velocity. Go back to either of your two oringinal equations and compute xf - xi. That is the distance.

7. Dec 1, 2006

### XPX1

but how do I find xf or xi if I know xi = 0 and xf = ?

8. Dec 1, 2006

### XPX1

I plugged it into the position versus time graph and got xf=14.663

xf=xi+vixt+.5at^2

xf=0+16(1.15)+.5(-5.65)(1.15)^2

9. Dec 1, 2006

### XPX1

Woohoo I got b. correct! But now for a. Fnet=MA then Fnet= 1020(-5.65)

Fnet = -5763

Then the average force exerted on the car while breaking is -5763 kN?

kN im guessing is kilonewtons?

10. Dec 1, 2006

### PhanthomJay

You are not keeping track of your units. 1Kg*m/s^2 = 1 newton

11. Dec 1, 2006

### XPX1

so how do I convert -5763 to kilonewtons?

Infact what is -5763? If the mass is kilograms and the acceleration was in.. meters?

Then.. -5763 would be kilograms right? 1 kilogram = 9.8 newtons? and 1000 newtons = 1 kilonewton? so -5763 would be -57630?

12. Dec 1, 2006

### PhanthomJay

You have to study up on the units. The acceleration is in meters/second^2. Velocity is in meters/second. Distance is in meters. And 1 newton is shorthand for 1Kg*m/second^2. and 1000N = 1kN .

F=ma
F=
1050Kg(-5.65m/s^2) = -5763Kg*m/s^2 = -5763N
-5763N(1kN/1000N) = - 5.763kN

As an aside, you should note that the fprmula for weight, which is unrelated to this problem, is W=mg, so 1Kg weighs (1)9.8 Kg(m)/s^2 = 9.8 Newtons

13. Dec 1, 2006

### XPX1

-5.763 kN was not correct, I still dont understand!

14. Dec 1, 2006

### PhanthomJay

Maybe the car was moving to the left. Try using +5.763kN.
EDIT: Oh, now i see the problem asked for the magnitude of the force, not the direction. The minus sign means that the force is opposite the direction of the car's motion. The magnitude of the force is 5.763kN.

Last edited: Dec 1, 2006
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