Posted by **illy** on Sunday, February 19, 2012 at 5:40pm.

At t=0 , a particle starts at the origin with a velocity of 6 feet per second and moves along the x-axis in such a way that at time t its acceleration is 12t^2 feet per second per second. Through how many feet does the particle move during the first 2 seconds?

- calculus -
**Reiny**, Sunday, February 19, 2012 at 5:52pm
a = 12t^2

velocity = 4t^3 + c

when t = 0, v = 6

6 = 4(0) + c

c = 6

v = 4t^3 + 6

distance = t^4 + 6t + k

when t = 0, distance = 0 ---> k = 0

distance = t^4 + 6t

so when t = 2

distance = 2^4 + 6(2) = 28 feet

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