What Is the Stopping Distance for Cars A and B?

In summary: The only energy we've talked about is gravitational potential energy = mghNot yet. The only energy we've talked about is gravitational potential energy = mghOkay, in this case the alternative method I was thinking about is not suitable. Just stick with the SUVAT method but remember that, given a choice of equations, you can pick whichever one you want in order to solve the problem.In summary, we discussed the calculation of stopping distance for two cars using the SUVAT equations. These equations involve five variables and five equations, each relating four of the variables. The final distance can be calculated using any of the equations, depending on the known variables. The alternative method using kinetic energy and force * distance was not suitable
  • #1
Barclay
208
1

Homework Statement


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 traveling at 30km/h and car B traveling 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.

Homework 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 Bspeed=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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  • #2
The speeds of the cars are not constant with time, but rather decreasing with time, therefore your calculations will not be correct.
 
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  • #3
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

Homework Statement

[/B]

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 traveling at 30km/h and car B traveling 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.

Homework Equations



F = ma

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

speed = distance / time = d/t

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

t=(v-u)/a

t = (12.5-0)/6.5

t= 1.92s to stop the car Bspeed=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
 
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  • #4
Fightfish said:
The speeds of the cars are not constant with time, but rather decreasing with time, therefore your calculations will not be correct.

Thanks. I've realized 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
Barclay said:
Thanks. I've realized 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
What about the SUVAT equations? Do you know what these are?
 
  • #6
SteamKing said:
What about the SUVAT equations? Do you know what these are?

No never heard of the SUVAT equation. Please tell me more. Thank you
 
  • #7
Barclay said:
No never heard of the SUVAT equation. Please tell me more. Thank you

Try these on for size:

https://sentynel.com/media/old/equations.html
 
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  • #8
Thanks SteamKing. I used the SUVAT equation and got the right answer with two different SUVAT formula.

Please note that the textbook 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 :) FANTASTICEQUATION 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
Barclay said:
perhaps there's a simpler method
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?
Barclay said:
Can you tell me which of the EQUATIONS 1 or 2 should I have used
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
haruspex said:
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?

Not yet
 

FAQ: What Is the Stopping Distance for Cars A and B?

1. What is the relationship between force, mass, and distance?

The relationship between force, mass, and distance is described by Newton's Second Law of Motion. This law states that the force applied to an object is equal to its mass multiplied by its acceleration, and the acceleration is directly proportional to the distance over which the force is applied. In simpler terms, the greater the force applied to an object, the greater its acceleration will be, and the greater the distance over which the force is applied, the greater the object's acceleration will be.

2. How do force, mass, and distance affect each other in an equation?

In an equation, force is represented by the letter F, mass by the letter m, and distance by the letter d. The relationship between these three variables can be expressed as F = m x a, where a is the acceleration. This equation shows that force is directly proportional to both mass and acceleration, and indirectly proportional to distance.

3. How does increasing the force affect the mass and distance of an object?

If the force applied to an object is increased, it will cause the object to accelerate at a greater rate. This means that the mass of the object will remain constant, while the distance it travels over a given time will increase. However, if the force is increased while the mass remains constant, the object will also experience a greater acceleration.

4. Can an object's mass affect the amount of force needed to move it a certain distance?

Yes, an object's mass does affect the amount of force needed to move it a certain distance. According to Newton's Second Law of Motion, the greater an object's mass, the greater the force needed to accelerate it over a given distance. This means that a heavier object will require more force to move it the same distance as a lighter object.

5. How does distance affect the force needed to move an object?

The distance over which a force is applied affects the amount of force needed to move an object because it is directly proportional to acceleration. This means that the greater the distance, the greater the acceleration of the object, and thus, the greater the force needed to move it. However, if the distance is decreased, the force needed to move the object will also decrease, but the acceleration may remain the same.

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