Displacement with magnitude of deceleration

1. Sep 25, 2007

Jacque77

You are driving your car and the traffic light ahead turns red. You apply the brakes for 3.59 s, and the velocity of the car decreases to +4.99 m/s. If the car's deceleration has a magnitude of 2.53 m/s2, what is the car's displacement during this time?

I know Δd=Vi(Δt)+1/2at2. I am not sure how to handle the deceleration magnitude and if this is the best formula to use.

2. Sep 25, 2007

learningphysics

You can use this formula (for some reason it is less commonly mentioned or used):

d = vf*t - (1/2)at^2

where vf is the final velocity. since it is decelerating a = -2.53m/s^2

or you can you use formula with vi... so first get vi using

vf = vi + at

3. Sep 25, 2007

Jacque77

so if I understand you properly:
d= 4.99m/s * 3.59s - (.5)(-2.53m/s^2)(3.59)^2

I get 17.914--16.30 = 34.21

Am I using the deceleration properly? I haven't had physics in 30+ yrs

4. Sep 25, 2007

learningphysics

Yup. That's right.

Also try calculating vi and then using

d = vi*t + 0.5at^2

you should get the same result that way also. that's a good way to double check your answer.

5. Sep 25, 2007

Jacque77

I don't understand how to use the second formula with the numbers given - the Vf=vi+at.

6. Sep 25, 2007

learningphysics

you can get vi from that equation:

4.99 = vi + (-2.53)(3.59)

vi = 14.0727

then you can use d = vi*t + (1/2)at^2

7. Sep 25, 2007

learningphysics

Then using

d = vi*t + (1/2)at^2 = 14.0727*3.59 + (1/2)(-2.53)(3.59)^2 = 34.2175m

So it's the same answer (only a slight difference due to rounding)

The best way to solve this problem, is using d = vf*t - (1/2)at^2 as you did... but for some strange reason this formula isn't mentioned in physics texts very often... that's why I mentioned this other method also...

8. Sep 25, 2007

Jacque77

OMG! It worked!! (14.0727)(3.59)+(.5)(-2.53)(3.59)^2 is
50.52-16.30 is 34.21!!!!!! yipee.