Displacement with magnitude of deceleration

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Homework Help Overview

The problem involves calculating the displacement of a car that is decelerating after applying brakes for a specific duration. The context includes the initial and final velocities, the magnitude of deceleration, and the time of deceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various formulas for calculating displacement, including those involving initial and final velocities. There is uncertainty about the correct application of deceleration and the formulas themselves.

Discussion Status

Some participants have provided alternative formulas and attempted calculations, while others express confusion about how to apply the equations correctly. There is a recognition that different approaches yield similar results, but no consensus on the preferred method has been established.

Contextual Notes

Participants mention a lack of recent physics knowledge and question how to handle the deceleration in their calculations. There is also a note that one of the formulas discussed is not commonly referenced in physics texts.

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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