What is the exact location of the object at time 3? Integrals question.?

An object originally located at x=4. Beginning at time zero, the object moves along the x axis with velocity given by -5-2t+4t^2. Determine the exact location of the object at time 3. What is the exact location of the object at time 3? Integrals question.?
displacement is integral of velocity





v(t) = -5 - 2t + 4t虏


s(t) = 鈭?5 - 2t + 4t虏 dt


= -5t - t虏 + 4t鲁/3 + C





Plug in known value


s(0) = 4


S(0) = -5(0) - (0)虏 + 4(0)鲁/3 + C


4 = -5(0) - (0)虏 + 4(0)鲁/3 + C


C = 4





s(t) = 4 - 5t - t虏 + 4t鲁/3





s(3) = 4 - 5(3) - (3)虏 + 4(3)鲁/3


= 4 - 15 - 9 + 36


= 16What is the exact location of the object at time 3? Integrals question.?
v(t) = -5 -2t + 4t^2 = dx/dt integrate dx from 4 to final position and the time function from 0 to 3. So,


INT{dx}from 4 to Xf = INT{-5 -2t + 4t^2} from 0 to 3


Xf - 4 = -15 -9 + (4/3)*27


Xf = -15 -9 + (4/3)*27 + 4


Xf = 16


So the object started at x = 4 and ended up at x = 16
v(t) = 4t^2 - 2t -5





Integrate to get:





x(t) = (4/3)t^3 -t^2 -5t + C.





C = x(0) = 4.





x(t) = (4/3)t^3 -t^2 -5t + 4.





x(3) = (4/3)(3)^3 -(3)^2 -5(3) + 4.





I'll let you do the rest.



plugging 3 into the equation you come up with x=25, so adding the original 4 to that x=29.
25 (make a graph)