# Force, mass, distance question

1. Dec 25, 2014

### Barclay

1. The problem statement, all variables and given/known data
Seasons greetings everyone. Merry Christmas and happy new year to all.

The brakes of a car can provide 6.5kN of force and the mass of the car is 1000kg. What is the stopping distance for car A travelling at 30km/h and car B travelling at 45km/h.

The answer in the book is A = 5.3m B = 12m

I did not get these answers and I cant figure out where I made the error. Please advise.

2. Relevant equations
f=ma
a=rate of change of speed = (v-u)/t
speed = distance / time = d/t

3. The attempt at a solution

For both cars f=ma
a=(6.5 x 103)/1000 = 6.5m/s2

Car A: 30km/h = 8.3m/s2

Car B : 45km/h = 12m/s2

Acceleration for car A: a=(v-u)/t
t=(v-u)/a t = (8.3-0)/6.5
t = 1.27s time to stop the car A

Acceleration for car B: a=(v-u)/t
t=(v-u)/a = (12-0)/6.5
t= 1.84s to stop the car B

speed=d/t so for car A distance = speed x time = 8.3 x 1.27 = 10.5m
speed=d/t so for car B distance = speed x time = 12 x 1.84 = 22m

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Last edited by a moderator: Dec 25, 2014
2. Dec 25, 2014

### Fightfish

The speeds of the cars are not constant with time, but rather decreasing with time, therefore your calculations will not be correct.

3. Dec 25, 2014

### Barclay

I made an error on my 1st post above .... I wrote the velocities as metres per second SQUARED accidentally and a few other errors

So I have re-posted the question and my attempt below

1. The problem statement, all variables and given/known data

The brakes of a car can provide 6.5kN of force and the mass of the car is 1000kg. What is the stopping distance for car A travelling at 30km/h and car B travelling at 45km/h.

The answer in the book is A = 5.3m B = 12m

I did not get these answers and I can't figure out where I made the error. Please advise.

2. Relevant equations

F = ma

a= rate of change of speed = (v-u)/t

speed = distance / time = d/t

3. The attempt at a solution
For both cars F = ma
a =(6.5 x 1000)/1000 = 6.5m/s2 = deceleration for each car

Car A: 30km/h = 8.3m/s = speed of car converted to m/s
Car B : 45km/h = 12.5m/s = speed of car converted to m/s

Acceleration for car A: a=(v-u)/t

So t=(v-u)/a

t = (8.3-0)/6.5

t = 1.28s time to stop the car A

Acceleration for car B: a=(v-u)/t

t=(v-u)/a

t = (12.5-0)/6.5

t= 1.92s to stop the car B

speed=d/t so for car A distance = speed x time = 8.3 x 1.28 = 10.6m
speed=d/t so for car B distance = speed x time = 12.5 x 1.92 = 24m

Last edited: Dec 25, 2014
4. Dec 25, 2014

### Barclay

Thanks. I've realised from what you said that the equation speed = distance /time will not apply (used in the last part of my calculation).

So what formula can I use? I've got TIME DECELERATION FORCE MASS. I need a formula that brings in DISTANCE. I'm stuck now

5. Dec 25, 2014

### SteamKing

Staff Emeritus
What about the SUVAT equations? Do you know what these are?

6. Dec 25, 2014

### Barclay

No never heard of the SUVAT equation. Please tell me more. Thank you

7. Dec 25, 2014

### SteamKing

Staff Emeritus
8. Dec 25, 2014

### Barclay

Thanks SteamKing. I used the SUVAT equation and got the right answer with two different SUVAT formula.

Please note that the text book I'm reading does not even mention this formula (yet) so perhaps there's a simpler method [THOUGH HAVE NO IDEA WHAT IT MAY BE]. (The answer page just says for car A time = 1.28 s so distance = 5.3m and for car B time = 1.92 s so distance = 12 m).
I'm not criticising you here at all (just informing you that this SUVAT is not mentioned yet) but your idea of SUVAT is fantastic for me :):)

I'm going to show you my calculations and please can you tell me which one is meant to be used although both give the correct answers.

Anyway .... for CAR A
s = distance ???
u = initial velocity = 8.3
v = final velocity = 0
a = acceleration = -6.5
t = time = 1.28

EQUATION 1. s = ut + (1/2)at2

s = 8.3*1.28 + 0.5*(-6.5)*1.282

s = 10.624 + (-5.324)

s = +5.3m CORRECT ANSWER :) FANTASTIC

EQUATION 2. s = vt - (1/2)at2

s = 0*1.28 - 0.5*(-6.5)*1.282

s = 0- (-5.324)

s = +5.3 m CORRECT ANSWER :) FANTASTIC

Can you tell me which of the EQUATIONS 1 or 2 should I have used. In this case answers were correct with both equations. Thanks

9. Dec 25, 2014

### haruspex

Not a simpler method, but perhaps one which uses equations you've already been taught. Do you know about kinetic energy and energy = force * distance?
There's no 'should' here. They're all valid and you can use whichever you need to get an answer. Some routes may be quicker than others.
Depending on what form of SUVAT you've been taught, there are five standard variables and five equations. Each equation relates a different four of the five variables. So, given three knowns and one you wish to find, pick the equation that involves those four.

10. Dec 25, 2014

Not yet