What is the car's acceleration just before emerging from the turn?

[mujadeo]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

heres what i know, but can't 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?

Please help!

 

[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.

 

[Doc Al]

(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?)

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?

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

Use the distance.

 

[mgb_phys]

Isn't the corner a red herring?

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

v =

 

[Doc Al]

Isn't the corner a red herring?

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

(And the OP had already calculated the distance.)