anonymous
  • 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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
my*
Asylum15
  • 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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
ok..
anonymous
  • anonymous
how exactly would i sub it in?
Asylum15
  • Asylum15
One sec :)
Asylum15
  • 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
  • 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
  • anonymous
:o
Asylum15
  • Asylum15
Lost?
anonymous
  • anonymous
pnaa get it now.. never thought of using brackets lol.. continue plz =P
Asylum15
  • 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
  • anonymous
thx man
Asylum15
  • Asylum15
Respect.
anonymous
  • anonymous
dude, how did u get 48.28 ?.. when i do the quadratic.. dont even get close to it ..

Looking for something else?

Not the answer you are looking for? Search for more explanations.