Question

A 1900kg car starts from rest and drives around a flat 56-m-diameter circular track. The forward force provided by the car's drive wheels is a constant 1000N.
a. What is the magnitude of the car's acceleration at t=13s?
b. What is the direction of the car's acceleration at t=13s? Give the direction as an angle from the r-axis.
c. If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?


Attempt at a solution:
a= sqrt(tangential acc. sq + radial acc. sq)
a(t)= F/m
a(r)= mv^2/r
here v= at (?)
But I got a huge acceleration for doing that and it's a wrong answer. Where did i go wrong?

Thanks in advance.

Answer 1

I am wondering what forward friction is. It is hard to assume it is zero, but that affects what the forward acceleration.

1000-frictionforward= mass*acc forward

otherwise, what you laid out is the correct procedure.

a= sqrt((netforce/mass)^2+ (at)^2/r)

Notice I disagree with your a(r), I don't see the need for mass to be there.

Answer 2

Thank you so so much. Well I got the first two answers correct now. But in c i think i should use friction force which should be rubber upon concrete? Co-efficient of kinetic friction for that is 0.8. (I need the kinetic friction co-efficient here right?) But i'm not sure how to do this part. What will the approach be?

Answer 3

You need the coefficient

Answer 4

dasfasf
s

Answer 5

suckyamudaah