Question

I have a calculus question about initial velocity, pic included.

Answer 1

The constant acceleration of a ball shot up from the ground is a = -32 ft/sec^2.

The acceleration is defined as the instant rate of change of the velocity:

`a=(dv)/(dt)`

Integrating this equation, we have:

`v=\int a\cdot dt`

Since the acceleration is a constant function:

`\begin{gathered} v=\int-32\cdot dt \\ \\ v=-32\int dt \\ \\ v=-32t+v_o \end{gathered}`

The initial velocity is 48 ft/s, thus

`v=48-32t`

The ball will stop and return to the ground when the velocity is 0:

`48-32t=0`

Solving for t:

`t=(48)/(32)=1.5`

Now we find the displacement function by integrating the velocity:

`d=\int(48-32t)dt`

Integrating:

`\begin{gathered} d=48t-(32t^2)/(2)+d_o \\ \\ d=48t-16t^2+d_o \end{gathered}`

The ball was shot from the ground, so do = 0:

`d=48t-16t^2`

When t = 0, the position is d = 0.

When t = 1.5 seconds, the position is:

`\begin{gathered} d=48\cdot1.5-16(1.5)^2 \\ \\ d=72-36 \\ \\ d=36 \end{gathered}`

The ball goes up to 36 feet