Asked by anon
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.
A. 3t^3+t^2+C
B. 3t^3+2t^2+C
C. 18t+C
D. 20t+C
My answer is B.
Answers
Answered by
jai
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
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
Answered by
Jiskha
I don't know how to do this
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