1. ## Physics problems

SOrry if Im in wrong section but I got problems with my physics homework here it it:

A car was moving at a constant velocity of 40 m s-1 .The driver saw an obstacle in front and he immediately stepped o the brake pedal.He managed to stop the car in 8 s.The distance of the obstacle from the car when the driver spotted it was 180 m.How far was the obstacle from the car after it stopped?

Thanks for helping ,I tried my best but I just cant figured it.

2. Inertia. problem solved

3. Originally Posted by Zuhrain
Inertia. problem solved
Elaborate?I cant seem to find a formula under tht topic.

4. Originally Posted by aweosmejules
Elaborate?I cant seem to find a formula under tht topic.
He's just trolling you.

5. Originally Posted by Astro.
He's just trolling you.
LOL Dafuq?

6. Originally Posted by aweosmejules
SOrry if Im in wrong section but I got problems with my physics homework here it it:

A car was moving at a constant velocity of 40 m s-1 .The driver saw an obstacle in front and he immediately stepped o the brake pedal.He managed to stop the car in 8 s.The distance of the obstacle from the car when the driver spotted it was 180 m.How far was the obstacle from the car after it stopped?

Thanks for helping ,I tried my best but I just cant figured it.

Formula for you, then just calculate the rate of the velocity, considering it is a vector. Also put in the multivariables, don't forget that.

7. Originally Posted by aweosmejules
SOrry if Im in wrong section but I got problems with my physics homework here it it:

A car was moving at a constant velocity of 40 m s-1 .The driver saw an obstacle in front and he immediately stepped o the brake pedal.He managed to stop the car in 8 s.The distance of the obstacle from the car when the driver spotted it was 180 m.How far was the obstacle from the car after it stopped?

Thanks for helping ,I tried my best but I just cant figured it.
It's SUVAT.

s = ? What we want to find out.
u = 40 ms^-1
v = 0 ms^-1
a = X We don't know and probably don't need to know.
t = 8 s

To find the distance it takes for the car to stop, use s = t(u+v)/2).

So,
s = 8 * (40/2)
s = 8 * 20
s = 160 m

To find the distance the car stops from the obstacle, simply take the distance it takes to stop the car away from the distance the car was from the obstacle originally.

180 - 160 = 20

So the car stopped 20m away from the obstacle.

8. ## The Following 2 Users Say Thank You to Unknown For This Useful Post:

aweosmejules (02-24-2015),Mokou-Sama (02-24-2015)

9. Originally Posted by Unknown

It's SUVAT.

s = ? What we want to find out.
u = 40 ms^-1
v = 0 ms^-1
a = X We don't know and probably don't need to know.
t = 8 s

To find the distance it takes for the car to stop, use s = t(u+v)/2).

So,
s = 8 * (40/2)
s = 8 * 20
s = 160 m

To find the distance the car stops from the obstacle, simply take the distance it takes to stop the car away from the distance the car was from the obstacle originally.

180 - 160 = 20

So the car stopped 20m away from the obstacle.
Thanks for the help but what the differences between s=ut+1/2at2 and s = t(u+v)/2 that you given?

- - - Updated - - -

Is it just the same but two different ways to get them?

Edit:I pretty understand that u=40ms^-1 because of constant velocity so the v=0ms^-1 because the car was stopped and the constant velocity too?

10. Originally Posted by aweosmejules
Thanks for the help but what the differences between s=ut+1/2at2 and s = t(u+v)/2 that you given?

- - - Updated - - -

Is it just the same but two different ways to get them?

Edit:I pretty understand that u=40ms^-1 because of constant velocity so the v=0ms^-1 because the car was stopped and the constant velocity too?
No problem.

To be able to use the formula;

s = ut + 1/2 * (at^2) ,

the acceleration of the car would need to be know. Obviously this could be calculated because acceleration is the change in velocity over time. So by using the formula;

a = (v - u) / t ,

we could find this out. So yeah that formula could be used easily but it does involve an additional calculation. Yeah it's just another way to do it.

You're told that the car is moving at a constant velocity of 40 ms^-1 and decelerates to a stop. This means that the car was originally moving at 40 ms^-1. Therefore this is the initial velocity, u. The car stops so you know that the final velocity has to be 0 ms^-1, so v = 0.

11. ## The Following User Says Thank You to Unknown For This Useful Post:

aweosmejules (02-24-2015)

12. Originally Posted by Unknown

No problem.

To be able to use the formula;

s = ut + 1/2 * (at^2) ,

the acceleration of the car would need to be know. Obviously this could be calculated because acceleration is the change in velocity over time. So by using the formula;

a = (v - u) / t ,

we could find this out. So yeah that formula could be used easily but it does involve an additional calculation. Yeah it's just another way to do it.

You're told that the car is moving at a constant velocity of 40 ms^-1 and decelerates to a stop. This means that the car was originally moving at 40 ms^-1. Therefore this is the initial velocity, u. The car stops so you know that the final velocity has to be 0 ms^-1, so v = 0.
Thx for the help +rep! @arunforce pls close this thread.

13. Originally Posted by aweosmejules
Thx for the help +rep! arunforce pls close this thread.
No problem. Glad I could assist ya.

14. ## The Following User Says Thank You to Unknown For This Useful Post:

aweosmejules (02-24-2015)

15. Alevel physics haha, .... I remember this SUVAT!! :0

16. Brilliant, I was gonna solve it but Unknown came first.

17. That's some kindergarten physic shit right there. Answer is obviously 666m away.. Gawsh.