Question

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

Answer 1

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.

Answer 2

thank you

Answer 3

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