Posted by **Dave** on Monday, February 23, 2009 at 6:41pm.

The acceleration of a particle at a time t moving along the x-axis is give by: a(t) = 4e^(2t). At the instant when t=0, the particle is at the point x=2, moving with velocity v(t)=-2. Find the position of the particle at t=1/2

if you could show me how to get that please

- math -
**bobpursley**, Monday, February 23, 2009 at 6:50pm
postition= integral v dt

v= int a dt

find v first, you know the constant of integration from at t=2

Then integrate again to get position.

- math -
**Reiny**, Monday, February 23, 2009 at 6:55pm
a(t) = 4e^(2t)

so v(t) = 2e^(2t) + C (I integrated, since dv/dt = a)

when t=0, v=-2

-2 = 2e^0 + c

c = -4 and then

v)t) = 2e^(2t) - 4

since ds/dt = v

s = e^(2t) - 4t + k

when t=0, s=2 (I assume that is what you meant by x=2)

2 = e^0 - 4(0) + k

k = 1

then s(t) = e^(2t) - 4t + 1

s(1/2) = e^1 - 2 + 1

= e - 1

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