Questions: Learning Outcome IC1: Problem 5 (1 point) Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving as given by the following data. a(t)=t-2 s(0)=-10 v(0)=6 Find a function describing the position of the particle. s(t)=

Solution Steps

To find the position function \( s(t) \) of the particle, we need to integrate the acceleration function \( a(t) \) to find the velocity function \( v(t) \), and then integrate \( v(t) \) to find the position function \( s(t) \). We will use the initial conditions \( v(0) = 6 \) and \( s(0) = -10 \) to determine the constants of integration.

To find the velocity function \( v(t) \), we integrate the acceleration function \( a(t) = t - 2 \):

\[ v(t) = \int (t - 2) \, dt = \frac{t^2}{2} - 2t + C_1 \]

Using the initial condition \( v(0) = 6 \):

\[ v(0) = \frac{0^2}{2} - 2(0) + C_1 = 6 \implies C_1 = 6 \]

Thus, the velocity function is:

\[ v(t) = \frac{t^2}{2} - 2t + 6 \]

Next, we find the position function \( s(t) \) by integrating the velocity function \( v(t) \):

\[ s(t) = \int \left( \frac{t^2}{2} - 2t + 6 \right) \, dt = \frac{t^3}{6} - t^2 + 6t + C_2 \]

Using the initial condition \( s(0) = -10 \):

\[ s(0) = \frac{0^3}{6} - 0^2 + 6(0) + C_2 = -10 \implies C_2 = -10 \]

Thus, the position function is:

\[ s(t) = \frac{t^3}{6} - t^2 + 6t - 10 \]

Final Answer