The velocity of a particle, in meters per second, is v(t) = -t^2 +5t +6, where s(6) = 75 a. Find the acceleration, a(t), of the particle at any time t and at t = 2 secounds. b. Find the displacement,

a. The acceleration of the particle can be found by taking the derivative of the velocity function: {eq}a(t) = v'(t) = -2t + 5 {/eq}. At t = 2 seconds, the acceleration is {eq}a(2) = -2(2) + 5 = 1 \text{ m/s}^2 {/eq}. b. To find the displacement of the particle at any time t, we need to take the antiderivative of the velocity function: {eq}s(t) = int v(t) dt = -frac{1}{3}t^3 + \frac{5}{2}t^2 + 6t + C {/eq}, where C is the constant of integration. We can find C by using the fact that s(6) = 75: {eq}75 = -frac{1}{3}(6)^3 + \frac{5}{2}(6)^2 + 6(6) + C {/eq}. Solving for C, we get C = -24. Therefore, the displacement function is {eq}s(t) = -frac{1}{3}t^3 + \frac{5}{2}t^2 + 6t - 24 {/eq}. c. The total distance traveled from t = 0 to t = 10 is given by the expression {eq}int_0^{10} |v(t)| dt {/eq}. Using the velocity function, we can see that the particle changes direction at t = 3 and t = 5. Therefore, we need to split up the integral into three parts: {eq}int_0^3 (-v(t)) dt + int_3^5 v(t) dt + int_5^{10} v(t) dt {/eq}. Plugging in the velocity function and simplifying, we get {eq}int_0^3 (t^2 - 5t - 6) dt + int_3^5 (-t^2 + 5t + 6) dt + int_5^{10} (-t^2 + 5t + 6) dt = 10 {/eq}. Therefore, the total distance traveled is 10 meters.