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- anonymous

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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- anonymous

- jamiebookeater

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- anonymous

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

- anonymous

my*

- 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)

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- anonymous

ok..

- anonymous

how exactly would i sub it in?

- Asylum15

One sec :)

- 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).

- 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

- anonymous

:o

- Asylum15

Lost?

- anonymous

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

- 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

- anonymous

thx man

- Asylum15

Respect.

- anonymous

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

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