Posted by **anon** on Saturday, March 5, 2011 at 10:04pm.

1. A body is moving along a straight line with a velocity which varies according to the equation v=9t^2+2t, where v is in feet per second and t is in seconds. Find the expression for the distance as a function of time.

A. 3t^3+t^2+C

B. 3t^3+2t^2+C

C. 18t+C

D. 20t+C

My answer is B.

- Calculus 12th grade. -
**jai**, Saturday, March 5, 2011 at 10:25pm
recall that velocity is the change in position over change in time, or:

v = dx/dt

where x=distance, and t=time

thus equating this to the given function,

dx/dt = 9t^2 + 2t

dx = (9t^2 + 2t)dt

integrating,

x - xo = 3t^3 + t^2 + C

if we assume that the initial displacement is zero or xo = 0, this becomes:

x = 3t^3 + t^2 + C

your answer is correct~! :D

- Calculus 12th grade. -
**Jiskha**, Wednesday, March 16, 2011 at 7:53am
I don't know how to do this

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