# Kinematics od deccelerating train

1. Aug 2, 2005

### frozen7

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. How far does the train move during 40-s time interval starting at the instant the brakes are applied?

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. Aug 2, 2005

### marlon

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

Last edited: Aug 2, 2005
3. Aug 2, 2005

### FredGarvin

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:

$$V^2 = {V_o}^2 + 2a(X - X_o)$$

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

Last edited: Aug 2, 2005
4. Aug 2, 2005

### Staff: Mentor

How about writing the kinematic equations?

acceleration, a is given by $a = \frac{dv}{dt}$

and speed, v is given by $v = \frac{dx}{dt}$

where x is the displacement or distance.

At what time does the train stop?

5. Aug 2, 2005

### frozen7

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

6. Aug 2, 2005

### marlon

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

marlon

7. Aug 2, 2005

### frozen7

after 20s,the displacement = 200, and it is stop moving,v=0
For the last 20s, the train continue moving in different direction since the acceleration is negative sign.
Is it right?

8. Aug 2, 2005

### marlon

yes, so what is the displacement after 40 seconds

marlon

9. Aug 2, 2005

### frozen7

displacement = 0 after 40s
False? Why?
I found it by subtitute u=20, t=40, a=-1 to s=ut+1/2at^2

10. Aug 2, 2005

### marlon

err you are right, i think

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. Aug 2, 2005

### frozen7

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

12. Aug 2, 2005

### FredGarvin

It is a bit of a trick. It got me at first (after I read the problem correctly). The train stops after only 20 seconds. I would say that, since the brakes being applied would not allow the train to go backwards, then the displacement after t=20 is zero. Therefore, 20 seconds is the only time you need to consider the train as moving.

13. Aug 2, 2005

### marlon

yes, you have been correct all the way

marlon

14. Aug 2, 2005

### frozen7

Ya,the BRAKES is the most important part in this question I think. Brakes wont decelerate.
Thanks anyway.

15. Aug 2, 2005

### Staff: Mentor

They key is the 'acceleration' as long as the train is in motion. Therefore, once the train stops, it is no longer accelerating.

16. Aug 2, 2005

### GCT

The "deceleration" pertains to the opposing frictional force, you should note the equations for frictional force.