Displacement with magnitude of deceleration

In summary, the car's displacement during the 3.59 seconds of deceleration is 34.21 meters. This can be calculated using the formula d = vf*t - (1/2)at^2 with vf = 4.99 m/s and a = -2.53 m/s^2. Another method is to first calculate the initial velocity (vi) using the equation vf = vi + at and then using the formula d = vi*t + (1/2)at^2. Both methods give the same result of 34.21 meters.
  • #1
Jacque77
14
0
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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  • #2
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
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
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.
 
  • #5
I don't understand how to use the second formula with the numbers given - the Vf=vi+at.
 
  • #6
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
 
  • #7
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
OMG! It worked! (14.0727)(3.59)+(.5)(-2.53)(3.59)^2 is
50.52-16.30 is 34.21! yipee.
 

FAQ: Displacement with magnitude of deceleration

What is displacement with magnitude of deceleration?

Displacement with magnitude of deceleration refers to the change in position of an object while it is slowing down.

How is displacement with magnitude of deceleration calculated?

Displacement with magnitude of deceleration can be calculated using the equation: d = (v0² - v²) / 2a, where d is displacement, v0 is initial velocity, v is final velocity, and a is deceleration.

Is displacement with magnitude of deceleration the same as distance traveled?

No, displacement with magnitude of deceleration refers to the change in position of an object, while distance traveled refers to the total length of the path traveled by the object.

Can displacement with magnitude of deceleration be negative?

Yes, displacement with magnitude of deceleration can be negative if the object is moving in the opposite direction of its initial velocity while decelerating.

How does displacement with magnitude of deceleration relate to acceleration?

Displacement with magnitude of deceleration is the opposite of displacement with magnitude of acceleration. While deceleration refers to a decrease in velocity, acceleration refers to an increase in velocity. Both involve a change in position of an object over time.

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