# An object travels North at 6 m/s for 4 s and then travels South at 3 m/s for 5 s. What are the object's average speed and velocity?

Apr 13, 2017

$\text{Speed} = \frac{4.3333333333 \overline{3} m}{s}$

$\text{Velocity} = \frac{1 m}{s}$

#### Explanation:

Calculating the average speed is very easy but I'm first doing average velocity because it requires more attention

$\textcolor{w h i t e}{\times \times \times \times \times \times} O$
$N \leftarrow \leftarrow \leftarrow \leftarrow A \rightarrow \rightarrow \rightarrow \rightarrow S$

Where A is the point where the object started from
N is north and S is south and is considered the midway between north andsouth. In our real earth that's equator
O is the object

The main trick here is to measure the distance between A and the point where the car stops.

Distance covered towards North

$\frac{6 m}{s} \times 4 = 24 m$

It is 24meters ahead of A towards north direction

It can be represented like this
$\textcolor{w h i t e}{\times \times \times \times} 24 m$
$\textcolor{w h i t e}{\times \times x} O \textcolor{w h i t e}{\times} \leftrightarrow$
$N \leftarrow \leftarrow \leftarrow \leftarrow A \rightarrow \rightarrow \rightarrow \rightarrow S$

Distance between A and O = 24m

Now calculate the distance moved towards South

$\frac{3 m}{s} \times 5 = 15 m$

$\textcolor{w h i t e}{\times \times x} O \rightarrow \text{by 15 meters}$
$N \leftarrow \leftarrow \leftarrow \leftarrow A \rightarrow \rightarrow \rightarrow \rightarrow S$

Distance moved from A

= 24m -15m = 9m

Time elapsed

4s + 5s = 9s

= 9m/9s = 1m/s is the velocity

$\text{Speed" = "distance"/"time}$

Distance is the distance covered can be anywhere in any direction in any zigzag pattern means the object can move back and front , eft and right.

So distance covered = 24m + 15m = 39m

Time elapsed

4s + 5s = 9s

= $\text{distance"/"time} = \frac{39 m}{9 s} = \frac{4.3333333333 \overline{3} m}{s}$