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## abz_tech 3 years ago A body leaves a point A and moves in a straight line with a constant velocity of 40 m/s. Ten seconds later another body which is at rest at A is given an acceleration of 2 m/s and moves in the same direction as the first body. How long does it take the second body to catch up with the first? How far from A does this occur?

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1. abz_tech

umm formulas u might need: v=u+at s=ut+1/2t^2 v^2=u^2 + 2as .. damn it can't get y head around the 10 sec part =/.. help would be really appreciated

2. abz_tech

my*

3. Asylum15

Write down what you know first: Body leaving Point A: v = 40 m/s a = 0 (since its got a constant velocity, it has no acceleration) Body 2 leaving Point A: u = 0 (initial velocity) a = 2 m/s t = 10 secs In the first part you need to find t (time) In the second part you need to find s (distance)

4. abz_tech

ok..

5. abz_tech

how exactly would i sub it in?

6. Asylum15

One sec :)

7. Asylum15

s = ut + ½ at² We'll let the time taken for the second body to catch the first be t. So, for the first body: s = 40(t + 10) (since this body has a 10-second start).

8. Asylum15

And, for the second body: s = ½(2)t² = t² When the second body catches the first these two distances are equal. So t² = 40t + 400 Or t² - 40t - 400 = 0

9. abz_tech

:o

10. Asylum15

Lost?

11. abz_tech

pnaa get it now.. never thought of using brackets lol.. continue plz =P

12. Asylum15

Well, you have a quadratic there. Which gives t = 48.2842712474619s (since t >0) Substitute this value back into the first equation for s to get s = 2331.3708499m To a sensible degree of accuracy t = 48.3s and s = 2330m

13. abz_tech

thx man

14. Asylum15

Respect.

15. abz_tech

dude, how did u get 48.28 ?.. when i do the quadratic.. dont even get close to it ..

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