How can I find the distance a car travels when the brakes are applied?

[BuBbLeS01]

Question:
A. A car that weighs 12000.0 N is initially moving at a speed of 57.0 km/hr when the brakes are applied and the car is brought to a stop in 2.6 s. Find the magnitude of the force that stops the car, assuming it is constant.

B. What distance does the car move during this time?

Okay so I drew the diagram and I am planning on using the f=ma equation but I think I need to solve for a first using Vf=Vo + at. Am I on the right track or am I completely off??

 

[learningphysics]

Yes, you're on the right track.

 

[BuBbLeS01]

Okay so I did...
(57 km/hr)^2 = a x (2.6s)
and I got 1249.6 m/s^2...which way wrong...where am I going wrong?

 

[BuBbLeS01]

Wait I wrote that backwards sorry...

I got...

0 = (57 km/hr)^2 +a(2.6s)

 

[learningphysics]

Wait I wrote that backwards sorry...

I got...

0 = (57 km/hr)^2 +a(2.6s)

Why are you squaring 57?

 

[PhanthomJay]

Why are you squaring the velocity? And you've got to be consistent with your units...you can't have seconds on one side of the equation and hours on the other without introducing a conversion factor.

 

[BuBbLeS01]

oh my gosh I don't know why I am squaring it...I am looking at the wrong equation. Okay I changed 57 km/hr to 0.01583 km/s. Now I have...
0 = 0.01583 km/s + a (2.6s)...
(-0.01583 km/s) / (2.6s) = a

so now I can use f=ma
f= (12000.0 N) x (0.0060897 km/s^2)

 

[learningphysics]

oh my gosh I don't know why I am squaring it...I am looking at the wrong equation. Okay I changed 57 km/hr to 0.01583 km/s. Now I have...
0 = 0.01583 km/s + a (2.6s)...
(-0.01583 km/s) / (2.6s) = a

so now I can use f=ma
f= (12000.0 N) x (0.0060897 km/s^2)

careful... you should use m/s for velocity and m/s^2 for acceleration (because your force equation requires kg for mass and m/s^2 for acceleration... to get the result in N)...

and also 12000.0N isn't the mass, it's the weight... get the mass of the object.

 

[BuBbLeS01]

careful... you should use m/s for velocity and m/s^2 for acceleration (because your force equation requires kg for mass and m/s^2 for acceleration... to get the result in N)...

and also 12000.0N isn't the mass, it's the weight... get the mass of the object.

okay so this is what I have now...

0=15.83 m/s + a x (2.6s)
(-15.83m/s) / (2.6s) = a = (-6.08846)

w=mg
12000/9.8=1224.49

f=ma
1224.49 * -6.08846 = -7455.26 N
That is really big!

 

[learningphysics]

okay so this is what I have now...

0=15.83 m/s + a x (2.6s)
(-15.83m/s) / (2.6s) = a = (-6.08846)

w=mg
12000/9.8=1224.49

f=ma
1224.49 * -6.08846 = -7455.26 N
That is really big!

Looks good. That's the right answer.

 

[BuBbLeS01]

Looks good. That's the right answer.

It says its incorrect...I don't know what is wrong.

 

[learningphysics]

It says its incorrect...I don't know what is wrong.

I don't think you carried enough decimal places... I get -7456.8N

 

[BuBbLeS01]

I don't think you carried enough decimal places... I get -7456.8N

should I be entering it as positive since its the magnitude?

 

[learningphysics]

should I be entering it as positive since its the magnitude?

Yes! I didn't notice that it asked for magnitude!

 

[BuBbLeS01]

Yes! I didn't notice that it asked for magnitude!

Yay it's right! Hehe...Thank you so much for your time! I really do appreciate it!

 

[learningphysics]

Yay it's right! Hehe...Thank you so much for your time! I really do appreciate it!

no prob.

 

[BuBbLeS01]

How do I start the second part of this question?

 

[PFStudent]

Hey,

If I remember right I believe you should use the displacement function,

$${x}(t) = {x}_{0} + {v}_{x_{0}}{t} + \frac{1}{2}{a}_{x}{t}^{2}$$

I think that should work.

Thanks,

-PFStudent

 

[BuBbLeS01]

Thats what I used but it says incorrect?
I put...

Xt = 0 + (57 km/hr) x (2.6s) + (1/2)(21.9km/s^2)(2.6s^2)

 

[learningphysics]

Thats what I used but it says incorrect?
I put...

Xt = 0 + (57 km/hr) x (2.6s) + (1/2)(21.9km/s^2)(2.6s^2)

don't mix up your units... do everything in m, s, m/s, m/s^2

 

[BuBbLeS01]

For the acceleration I have a formula of (Vf-Vo)/t...that doesn't seem right...is it?

 

[learningphysics]

For the acceleration I have a formula of (Vf-Vo)/t...that doesn't seem right...is it?

It is right, but you need to use the right units... you can't use km/hr then use seconds for time... use m/s for velocity... convert the velocity to m/s.

units must be consistent.

 

[BuBbLeS01]

Oops! here we go..

Xt = 0 + (15.83 m/s) x (2.6s) + (1/2)(7.92m/s^2)(2.6s^2)

and its still saying my answer is wrong :(

 

[learningphysics]

Oops! here we go..

Xt = 0 + (15.83 m/s) x (2.6s) + (1/2)(7.92m/s^2)(2.6s^2)

and its still saying my answer is wrong :(

where are you getting the 7.92? 7.92 is not 15.83/2.6. Also acceleration is negative.

 

[BuBbLeS01]

Oh my gosh I divided by 2 instead of 2.6! My answer is still wrong though. I got 33.24 mXt = 0 + (15.83 m/s) x (2.6s) + (1/2)(-6.088m/s^2)(2.6s^2)

 

[learningphysics]

Oh my gosh I divided by 2 instead of 2.6! My answer is still wrong though. I got 33.24 m


Xt = 0 + (15.83 m/s) x (2.6s) + (1/2)(-6.088m/s^2)(2.6s^2)

you didn't do (2.6)^2... it's (1/2)at^2...

There's another formula you can also use to get the distance... it doesn't have acceleration in it. Check your answer both ways and compare them...