• May 11th 2010, 08:25 AM
coolhacker
Hi

everyone

i got clue how to do it?

here it is:

1. if v=5+24t - 3t^2 , where v m/s is the velocity of the body at a time t seconds.

Determine:
(i) the expression for distance, S, in metres, m.
(ii) how far it moves in the interval t=1 sec to t=5 sec.

thxxxxxxx
(Nod)
• May 11th 2010, 09:52 AM
sa-ri-ga-ma
v=5+24t - 3t^2

v = ds/dt =5+24t - 3t^2

So ds = (5+24t - 3t^2)dt

$s = \int{(5+24t - 3t^2)}dt$

Find the integration to find the expression for s.
• May 11th 2010, 10:10 AM
coolhacker
how do i do this the integration to find the expression for s plz can you help ?
thxx
• May 11th 2010, 06:08 PM
sa-ri-ga-ma
Quote:

Originally Posted by coolhacker
how do i do this the integration to find the expression for s plz can you help ?
thxx

Have you studied integration?
• May 12th 2010, 09:29 AM
coolhacker
no not yet! my teacher gives me questions tht i havnt studyed so needed help
• May 12th 2010, 06:09 PM
sa-ri-ga-ma
http://www.mathhelpforum.com/math-he...6feaab99-1.gif

s = 5t + 24t^2/2 - 3t^3/3

For displacement, find s1 and s2 for two time interval and then s = s2 - s1.
• May 13th 2010, 09:45 AM
coolhacker
ok thxx for the help i have got the answer

thxx