Question

The position equation for the movement of a particle is given by s = (t²-1)³ when s is measured in feet and t is measured

in seconds. Find the acceleration at t = 2 seconds.

Answer 1

To find the acceleration at t = 2 seconds, differentiate the position equation with respect to time to find the acceleration function and substitute t = 2 into the acceleration function.

Step-by-step explanation:

To find the acceleration at t = 2 seconds, we need to differentiate the position equation with respect to time. The given position equation is s = (t²-1)³. Differentiating this equation will give us the velocity function, and differentiating the velocity function will give us the acceleration function.

Let's first differentiate the position equation to find the velocity function:

ds/dt = 3(t²-1)²(2t)

Now, let's differentiate the velocity function to find the acceleration function:

d²s/dt² = 6(t²-1)(3(t²-1)-1)

To find the acceleration at t = 2 seconds, we can substitute t = 2 into the acceleration function:

d²s/dt² = 6(2²-1)(3(2²-1)-1)

Simplifying the equation will give us the acceleration at t = 2 seconds.

Answer 2

The acceleration at t = 2 seconds is 18 ft/s².

Step-by-step explanation:

The problem involves finding the acceleration of a particle at a specific moment given its position equation s = (t²-1)³. To find acceleration, we need to differentiate the position equation twice with respect to time, because acceleration is the second derivative of position with respect to time. First, we find the velocity v(t) by differentiating position s(t) with respect to time t, then we differentiate the velocity to get acceleration a(t).

The position equation for the movement of a particle is given by s = (t²-1)³ when s is measured in feet and t is measured in seconds. To find the acceleration at t = 2 seconds, we need to find the second derivative of the position equation. Taking the derivative gives us a = 6t(t²-1). Substituting t = 2 into the equation gives us a = 6(2)(2²-1) = 18 ft/s².