# Car going round corner

1. Jul 23, 2007

1. The problem statement, all variables and given/known data

A car moving at a speed of 27 m/s enters a curve that describes a quarter turn of radius 126 m. The driver gently applies the brakes, giving a constant tangential deceleration of magnitude 1.2 m/s2.

2. Relevant equations

a) Just before emerging from the turn, what is the magnitude of the car's acceleration?

3. The attempt at a solution

heres what i know, but cant seem to put it all together for this problem.
v-initial = 27m/s
v-final (as it emerges from the turn) = ?
a = 1.2 (the radial acceleration component has to be zero because it is going round an arc right? (pie/2 .. also s = 197.9m )
(the tangetial acceleration is the one that changes, correct?)

I need to find v-final but how can i do that without the time????

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 23, 2007

### mgb_phys

You can re-arange s = u t + 0.5 a t^2 to give
v^2 = u^2 + 2 a S (Sorry for not putting it into latex)

Good start is to write down all the numbers you know, or can easily work out.

3. Jul 23, 2007

### Staff: Mentor

Not correct--just the opposite! The tangential acceleration is given as constant--the tangential speed changes, of course. The radial acceleration is not zero! Hint: How do you calculate centripetal acceleration?

Use the distance.

4. Jul 23, 2007

### mgb_phys

Isn't the corner a red herring?

Distance = 0.2 * 2pi*126 m
U = 27
a = 1.2 m/s^2

v =

Last edited: Jul 23, 2007
5. Jul 23, 2007

### Staff: Mentor

Not at all. Note that they take care to specify "Just before emerging from the turn..."

(And the OP had already calculated the distance.)