Question 1196122
The acceleration function is {{{a(t)=2t-6}}}. By definition, acceleration is the rate of change in velocity, aka the derivative of velocity. So taking the integral gives us the velocity as {{{v(t)=t^2-6t+c}}}, for some constant c. We know the initial velocity is 2 m/s, aka v(0)=2. This means that c=2, so {{{v(t)=t^2-6t+2}}}. This allows us to answer a), {{{v(5)=5^2-6*5+2=25-30+2=-3}}} meters per second. The velocity is the rate of change in position, aka the velocity is the derivative of the position. Taking the integral again gives us {{{s(t)=t^3/3-3t^2+2t+d}}}, where d is a constant. Since we're only considering displacement, we can assume the particle starts at position 0, so {{{s(t)=t^3/3-3t^2+2t}}}. We can now answer a), getting {{{s(5)=5^3/3-3*5^2+2*5=-70/3}}} meters of displacement.