Question

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?

Answer 1

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

Answer 2

my*

Answer 3

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)

Answer 4

ok..

Answer 5

how exactly would i sub it in?

Answer 6

One sec :)

Answer 7

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).

Answer 8

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

Answer 9

:o

Answer 10

Lost?

Answer 11

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

Answer 12

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

Answer 13

thx man

Answer 14

Respect.

Answer 15

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