Question

An rock's acceleration at time t is given by a(t) = 16t, and its initial velocity is 35. Find the velocity function v(t).

Answer 1

`\displaystyle v(t) = 8\, t^(2) + 35`.

Step-by-step explanation:

This question provided the following information:

• Acceleration of the object as a function of time, and
• The initial value of velocity (at time
  `t = 0`.)

The velocity function of this object can be found in the following steps:

• Integrate the acceleration function indefinitely with respect to time to obtain an expression of velocity with respect to time, plus a constant of integration that needs to be found.
• Make use of the initial value of velocity
  `v(0) = 35` to find the value of the constant of integration.

Using the power rule of integration, integrate the acceleration function with respect to time:

`\begin{aligned}v(t) &= \int a(t)\, d t \\ &= \int 16\, t\, d t \\ &= 16\, \left((1)/(2)\, t^(2)\right) + C \\ &= 8\, t^(2) + C\end{aligned}`.

In this expression,
`C` is the constant of integration that needs to be found.

To find the constant of integration
`C`, substitute the initial value
`v(0) = 35` and
`t = 0` into the expression
`v(t) = 8\, t^(2) + C`. Solve for
`C` as the unknown:

`v(t) = 8\, t^(2) + C`.

`v(0) = 8\, \left(0^(2)\right) + C`.

`35 = 8\, (0^(2)) + C`.

`C = 35`.

Hence, the velocity function would be:

`v(t) = 8\, t^(2) + 35`.