Questions: A particle moves along the x-axis. Its position is given by p(t)=2t^3-9t^2+15 inches where time is measured in seconds. What is the object's acceleration at t=5 seconds?

Solution Steps

To find the acceleration, we first need to determine the velocity function. The velocity is the first derivative of the position function \( p(t) \).

Given: \[ p(t) = 2t^3 - 9t^2 + 15 \]

The velocity function \( v(t) \) is: \[ v(t) = \frac{d}{dt} [2t^3 - 9t^2 + 15] \] \[ v(t) = 6t^2 - 18t \]

Next, we find the acceleration function by taking the derivative of the velocity function \( v(t) \).

The acceleration function \( a(t) \) is: \[ a(t) = \frac{d}{dt} [6t^2 - 18t] \] \[ a(t) = 12t - 18 \]

Now, we substitute \( t = 5 \) into the acceleration function to find the acceleration at that specific time.

\[ a(5) = 12(5) - 18 \] \[ a(5) = 60 - 18 \] \[ a(5) = 42 \]

Final Answer