Q. 81 - Calculus

Question

On earth, A falling object has a downward acceleration of 32 feet per second per second due to gravity. Suppose an object falls from an initial height of  $s_0$,With an initial velocity of $v_0$ feet per second, Use antiderivatives to show that the equations for the position and velocity of the object after t seconds are respectively $s\left(t\right)=-16t^2+v_{0}t+s^0$ and $v\left(t\right)=-32t+v_o\left(t\right)$

Step-by-Step Solution

Answer

We use antiderivatives to show that the equations for the position and velocity of the object after t seconds are respectively $$\begin{gathered} s\left(t\right)=-16t^2+v_{0}t+s_0 and \\ v\left(t\right)=-32t+v_0 \end{gathered}$$

Step 1: Given information

The downward acceleration of a falling object is 32 feet

Step 2: use antiderivatives to find the equations of v 0 , s 0

As the downward acceleration of the falling object is 32 feet per second second

We get,

$a\left(t\right)=-32$

Now when we derivate the velocity function we get acceleration

We use targeted guess nd check method

Since on differentiating the power function decreases its power by one Hence we start by function

$v\left(t\right)=t+v_0$ as at t=0 we the velocity becomes initial velocity

The equation is nearly equal now we just have to adjust the coefficients

We get,

$v\left(t\right)=-32t+v_0$        (1)

Step 3: Now we find equation of distance

Similarly,

We use targeted guess nd check method

Since on differentiating the power function decreases its power by one Hence we start by function

$s\left(t\right)=t^2+v_{0}t+s_0$

Which is nearly equal now we adjust the coefficients

$s\left(t\right)=-16t^2+v_{0}t+s_0$