# I need help with circular motion?

I was working on my physics homework and the circular motion stuff just confuses me. Can someone help me with this problem so that I can hopefully learn how to do the others on my own?

A race car starts from rest and accelerates at a uniform rate to a speed of 44.4 m/s in 12.8 s, moving on a circular track of radius 523 m. Calculate the total acceleration on the car the instant it is traveling at 4 m/s.

I know the answer should be approximately 3.171, but I have no idea how to arrive at that answer

### 2 Answers

- Andrew SmithLv 71 month agoFavorite Answer
To keep the car in a circle at any given speed we need a centripetal acceleration of v^2 / r = 4^2/523

To accelerate the car along a path we require an acceleration = dv/dt

= (44.4-0)/12.8

As both accelerations are perpendicular to each other the resultant lies on some angle between them and has a value calculated with Pythagoras' theorem.

ie a = sqrt( ac^2 + at^2)

= sqrt( (16/523)^2 + ( 44.4/12.8)^2)

Now I do not get the answer as 3.171 m/s^2 Although it is in the same general magnitude.

My answer is about 3.47 m/s^2 with the numbers you give me.

- Anonymous1 month ago
3.171 as an answer in an exam (even if correct, which it isn’t) will probably lose you 2 marks because:

a) units are missing;

b) the answer needs rounding to 3 sig. figs. to match the data.

__________

The car has 2 types of acceleration:

a) tangential (linear) acceleration (aₜ), which is the rate of increase of speed in the direction of velocity (this direction is a tangent tot he circular path at any instant);

b) radial (centripetal) acceleration (aᵣ) which accounts for the velocity’s changing direction as the car follows a circular path (aᵣ ‘s direction at any instant is radially inwards, towards centre of circle).

These 2 accelerations are perpendicular so get adde using vector addition.

_____________

a) Tangential acceleration has magnitude which is the rate of change of speed (same as something moving in a straight line.

aₜ = Δ|v|/t = (44.4 – 0)/12.8 = 3.469 m/s²

Note this is bigger than your official answer and we haven't even added-in aᵣ yet. So there is something wrong with the question/answer.

_____________

b) The radial acceleration is give by aᵣ = v²/r (or aᵣ = ω²r if more convenient). So when v= 4m/s:

aᵣ = v²/r = 4²/523 = 0.0306 m/s²

aᵣ is very small compared to aₜ, so to a good approximation can be ignored. But, if required, to add aₜ and aᵣ you do this:

a = √(aₜ² + aᵣ²) = √(3.469² + 0.0306²) = 3.47m/s² to 3 sig. figs. (same as aₜ)