Question

A bird is flying due east. Its distance from a tall building is given by

$x\left(t\right)=28.0m+\left(12.4m/s\right)t-\left(0.0450 m/s^3\right)t^3$.What is the instantaneous velocity of the bird when t = 8.00 s?

Short answer

The Instantaneous velocity of the bird at $t=8.00 s$ is$v_{8.00}=3.76m/s$

Step-by-step solution

Identification of the given data

The distance function for the bird is,

$x\left(t\right)=28.0m+\left(12.4m/s\right)\times t-\left(0.0450m/s^3\right)\times t^3$

and the given time is$t=8.00s$

Calculation of the instantaneous velocity of the bird

The instantaneous velocity can be expressed as,

$v_i=\frac{ds}{dt}$

Where $v_i$is the instantaneous velocity, is the displacement, and istime.

The expression for the distance is given as,

$x\left(t\right)=28.0m+\left(12.4 m/s\right)t-\left(0.0450m/s^3\right)t^3$

Instantaneous velocity expressed as

$$\begin{gathered} v_i=\frac{d}{dt}\left(28.0 m+\left(12.4m/s\right)t-\left(0.0450m/s^3\right)t^3\right) \\ =12.4m/s-\left(0.135m/s^3\right)\times t^2 \end{gathered}$$

Now,the substitution of $t=8.00s$in above expression will result in,

$\begin{array}{rcl}{v}_{8.00}& =& 12.4m/s-\left(0.135 m/s^3\right)\times {\left(8.00s\right)}^2\\ & =& 3.76m\end{array}$

Thus, the instantaneous velocity is $v_{8.00}=3.76 m/s$