- #1

Mogdan

- 10

- 0

## Homework Statement

u=60km/hr

m=1400kg

Friction of road (wet road)=0.4

angle of depression=3.65 degrees

radius of circle=156.3m

Distance of total circumference=983m

Distance of circumference needed to go=154m

Therefore it must go 56.4/360 degrees to go that distance.

[/B]

My problem is to find the acceleration of a car going 60km/hr, with a mass of 1400kg, going down a hill on an angle of depression of 3.65 degrees. It is turning towards the right and the radius of the circle is 156.3m. I am confused on how to calculate this as I have come up with 2 formulas to use but I don't know how to use these formulas together to find the overall acceleration of the car coming down the hill.

For the first formula I don't understand by it is negative acceleration if the car is going down a hill. Would I just say it is accelerating at that value instead of decelerating.

## Homework Equations

The first eformula is a=g(sin(theta)-Mu(cos(theta))

The second formula is a=(v^2/r) x (theta/360)

## The Attempt at a Solution

For the acceleration on a declined plane I used the first formula and subbed in the variables known

a=9.8(sin(3.65)-0.4(cos(3.65))

a=-3.288m/s

Second formula is

a=v^2/r x (theta/360)

=60^2/156.3 x (56.4/360)

a=3.6m/s

Last edited: