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# Speed = d/t formula

1. Just wondering why the s=d/t formula can't be used for part 2 of this question ?

The time in the first part was 4.16

I got the time, and speed, do distance= 5*4.16 = 20.8 , I know I could use suvat but not sure why this method couldn't be used as I get it wrong.
2. (Original post by Tulian)
Just wondering why the s=d/t formula can't be used for part 2 of this question ?

The time in the first part was 4.16

I got the time, and speed, do distance= 5*4.16 = 20.8 , I know I could use suvat but not sure why this method couldn't be used as I get it wrong.
s=d/t is for when you have constant speed (or average speed).

In this situation you have the speed at the start of the motion, but it's changing as the object is decelerating.
3. (Original post by Tulian)
Just wondering why the s=d/t formula can't be used for part 2 of this question ?

The time in the first part was 4.16

I got the time, and speed, do distance= 5*4.16 = 20.8 , I know I could use suvat but not sure why this method couldn't be used as I get it wrong.
s = d/t can only be used for objects travelling at a constant speed. The block B in this case is decelerating.

Use s = ut + 0.5 at^2 where:

s = ?
u = 5
a = -1.2
t = 4.17

(before anyone points it out, im just too lazy to type in the units above)
4. Thanks alot
5. (Original post by Tulian)
x
Was too late to help out, but I just wanted to wish you luck for your exam.

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