Q8E - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

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

Step 1: Identification of the given data

The distance function for the bird is, 

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

and the given time is $t=8.00s$

Step 1: 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,

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

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