**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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ANSWER

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**