What is the acceleration of a particle whose position is given by the equation s(t) = t^2 + 4t + 4 at the time t = 4?

To determine the acceleration of a particle moving along a straight line, we first need to understand the relationship between position, velocity, and acceleration.

The position of the particle is given by the formula:

s(t) = t2 + 4t + 4

First, we need to find the velocity of the particle, which is the first derivative of the position function with respect to time:

v(t) = ds/dt

Taking the derivative of s(t):

v(t) = d/dt(t2 + 4t + 4) = 2t + 4

Now, we can find the acceleration of the particle, which is the second derivative of the position function or the first derivative of the velocity function:

a(t) = dv/dt

Taking the derivative of v(t):

a(t) = d/dt(2t + 4) = 2

Notice that the acceleration is a constant value of 2, which means that it does not depend on the time t.

Now, let’s find the acceleration specifically when t = 4:

a(4) = 2

Thus, the acceleration of the particle when t = 4 is:

2 units per time squared.

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