Displacement with magnitude of deceleration

AI Thread Summary
The discussion focuses on calculating the displacement of a car that decelerates to a final velocity of +4.99 m/s over 3.59 seconds with a deceleration of 2.53 m/s². Two formulas for displacement are presented: d = vf*t - (1/2)at² and d = vi*t + (1/2)at². The first formula is confirmed as effective, while the second requires calculating the initial velocity (vi) using vf = vi + at. Both methods yield a similar displacement result of approximately 34.21 meters, demonstrating the consistency of the calculations. The conversation emphasizes the utility of both approaches in solving physics problems.
Jacque77
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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.
 
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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
 
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
 
Jacque77 said:
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

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.
 
I don't understand how to use the second formula with the numbers given - the Vf=vi+at.
 
Jacque77 said:
I don't understand how to use the second formula with the numbers given - the Vf=vi+at.

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