## Velocity Problems Calculus I need help with a calculus problem on accleration, velocity, and position?

Ok so the problem is
The acceleration of a particle at time t moving alon the x-axis is given by: a = 4e^(2t). At the instant when t = 0, the particle is at the point x = 2 moving with velocity v = -2. The position of the particle at t = 1/2 is ?
I am really confused with this problem. I don’t even know where to start. Thank you for your help!

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x = e – 1
or
x ≈ 1.72

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Given:
a = 4e^(2t)
When t=0 the velocity is v= -2
When t=0 the position is x=2

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NOTE:
Velocity is the integral of acceleration.
Position is the integral of velocity.
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a = 4e^(2t)
So,
V(t) = ⌠4e^(2t) dt
…… = 4⌠e^(2t) dt
…… = 4[½(e^(2t)] + C
…… = 2e^(2t) + C
Now, to determine C, we use the given information
When t=0 the velocity is v= -2
So,
V(0) = 2e^(2•0) + C
… -2 = 2eº + C
…. C = -4
Therefore,
V(t) = 2e^(2t) – 4
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Let S(t) be the position function. Then,
S(t) = ⌠2e^(2t) – 4 dt
…… = ⌠2e^(2t) dt – 4⌠dt
…… = 2[½e^(2t)] – 4t + C
…… = e^(2t) – 4t + C
Now, to determine C, we use the given information
When t=0 the position is x=2
So,
S(0) = e^(2•0) – 4(0) + C
…. 2 = 1 – 0 + C
…. C = 1
Therefore,
S(t) = e^(2t) – 4t + 1

You are to find the position when t = ½.
S(½) = e^(2•½) – 4(½) + 1
……. = e – 2 + 1
……. = e – 1
……. ≈ 1.72

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Calculus – Position, Velocity and Acceleration