OpenStudy (anonymous):

The acceleration of an object is given by a(t)= -32ft/s^2 . Find both the velocity function and position function if the initial velocity v(0)= 25ft/s and the initial position is given by s(0)= 40ft . Then find the position of the object at 2 seconds

5 years ago
OpenStudy (anonymous):

If you are taking Calculus based Physics, simply take the integral of acceleration twice. Once to find the velocity function and again to find the position function. \[v(t) = \int\limits a(t) dt = a(t) \cdot t + v(0)\]\[s(t) = \int\limits v(t) dt = {1 \over 2} a(t) \cdot t^2 + v(0) \cdot t + s(0)\] If you are not in Calculus based Physics, we first note that the acceleration is constant, that is it doesn't depend on time. From our constant acceleration equations of motion, we know that\[v(t) = a(t) \cdot t + v(0)\]\[s(t) = s(0) + v(t) \cdot t + {1 \over 2} a \cdot t^2\] Notice, how integration yields the same answer.

5 years ago
OpenStudy (anonymous):

thank you

5 years ago
OpenStudy (anonymous):

we can also do it by first finding the Final Velocity using eqn of motion Vf=Vi + at then the position can be calculate by another eqn of motion 2aS=Vf^2 - Vi^2

5 years ago