is there anything missing from this motion equation?

[inquisitive]

hello
have found this interesting question (interesting to me).... along with an equation that is supposed to allow the answer..... but I plug in the data and never get the same result as is correct.

 

could someone set it out with the data so it flows to give -690N (if correct).... thanks

(not an exam question or anything.... I found it and would like it as a model to find average brake force.... cheers)

 

 

A 1100-kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m. What is the average braking force acting on the car?
answer: (a) -690 N (b) -340 N © -410 N (d) -550 N

 

the answer is said to be (a) -690N
and the suggested equations are these:

 

Use Newton's second law in the form:
F = ma

 

And this equation of motion:
v2 = u2 + 2as

 

Where:
m = mass
F = braking force
v = final velocity
u = initial velocity
s = displacement

 

You end up with an equation like this:
F = (mv2)/2s

 

and the answer is -690 N

 

................................
appreciate thoughts, thanks

........ Have accomplished it.... (slightly obvious really)

 

didn't bother with the end equation...... simply used first 2 equations in own right:

* started with v2=u2+2as

* then plugged that into F=ma

* this gave answer in Newtons

 

All done

[swansont]

That's what I get, to two significant digits.

[Enthalpy]

As you have a brake distance, you could distribute the kinetic energy over it and get a force.

Believe it or not, I get 694N too.

[Mellinia]

You could also use the work-kinetic energy theorem and skip all those steps.
E_ktotal = W_total

1/2 m (v^2-u^2) = FS