- #1

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After 40s, displacement become zero,but the given answer is 200m. So, does it mean the 40s is just to confuse us? Does this question mean the train stop deceleration after it reach zero final velocity?

Thanks.

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- Thread starter frozen7
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- #1

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After 40s, displacement become zero,but the given answer is 200m. So, does it mean the 40s is just to confuse us? Does this question mean the train stop deceleration after it reach zero final velocity?

Thanks.

- #2

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no confusion going on...sorry

you used these formula's ???

v = 20 -t, velocity

x = x_o +20t - t²/2, position (the - just denotes the direction of the vectors at hand, it does not have an influence on the magnitude of the x here ; x_o is the initial position)

marlon

you used these formula's ???

v = 20 -t, velocity

x = x_o +20t - t²/2, position (the - just denotes the direction of the vectors at hand, it does not have an influence on the magnitude of the x here ; x_o is the initial position)

marlon

Last edited:

- #3

FredGarvin

Science Advisor

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No. The displacement does not become zero after 40 sec. The final speed is zero after 40 seconds.

Might I recommend the equation for constant acceleration that looks like:

[tex]V^2 = {V_o}^2 + 2a(X - X_o)[/tex]

EDIT: Crud. I completely misread the question....again. I need glasses.

Might I recommend the equation for constant acceleration that looks like:

[tex]V^2 = {V_o}^2 + 2a(X - X_o)[/tex]

EDIT: Crud. I completely misread the question....again. I need glasses.

Last edited:

- #4

Astronuc

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acceleration, a is given by [itex]a = \frac{dv}{dt} [/itex]

and speed, v is given by [itex]v = \frac{dx}{dt} [/itex]

where x is the displacement or distance.

At what time does the train stop?

- #5

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How if I use s = ut + 1/2 at^2 ? Isn`t the displacement = 0 by using this equation?

- #6

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frozen7 said:How if I use s = ut + 1/2 at^2 ? Isn`t the displacement = 0 by using this equation?

? If the displacement after 20 sec is non zero, then how it be after 40 sec ?

marlon

- #7

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For the last 20s, the train continue moving in different direction since the acceleration is negative sign.

Is it right?

- #8

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frozen7 said:

For the last 20s, the train continue moving in different direction since the acceleration is negative sign.

Is it right?

yes, so what is the displacement after 40 seconds

marlon

- #9

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displacement = 0 after 40s

False? Why?

I found it by subtitute u=20, t=40, a=-1 to s=ut+1/2at^2

False? Why?

I found it by subtitute u=20, t=40, a=-1 to s=ut+1/2at^2

- #10

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err you are right, i thinkfrozen7 said:displacement = 0 after 40s

False? Why?

I found it by subtitute u=20, t=40, a=-1 to s=ut+1/2at^2

i was under the impression that the car just moved backwards after the first 200. Because the v = 0 but the acceleration remains -1. But now i am readin that the acceleration = -1 is only to be incorporated when the breaks are applied.

I am confused now

marlon

- #11

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So....the time 40s is just to confuse us ,right?

- #12

FredGarvin

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- #13

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yes, you have been correct all the wayfrozen7 said:So....the time 40s is just to confuse us ,right?

marlon

- #14

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Thanks anyway.

- #15

Astronuc

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They key is the 'acceleration' as long as the train is in motion. Therefore, once the train stops, it is no longer accelerating.frozen7 said:A train is travelling down a straight track at 20m/s when the engineer applies the brakes,resulting in an acceleration of -1.0m/s^2 as long as the train is in motion.

- #16

GCT

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