A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive x axis point east.

 1. A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive x axis point east.

The data supplied by the problem is as follows. It is important to bear in mind that the time is given in units of minutes, therefore it is necessary to change the units to seconds:

• First displacement to the South (negative y-axis):
  
  $v_1=20.0m/s$
  
  $t_1=3.00min\ \Rightarrow180s$
  
  
  
• Second displacement west:
  
  $v_2=25.0m/s$
  

  
  $t_2=2.00min\ \Rightarrow180s$
  
  
• Third displacement to the Northeast:
  
  $v_3=30.0m/s$
  
  
  $t_3=1.00min\ \Rightarrow60s$

With the data provided above it is possible to determine:

(a) The vector resulting from the displacement

• First displacement to the South 
  
  

$v_1=20.0m/s $ 

$t_1=3.00min\ \Rightarrow180s$

Then the distance traveled during these 180s towards the South, is:

Thus, the vector of this first displacement is given by:

${\overline{r}}_1=-3600mj$

• Second displacement west:
  

$v_1=25.0m/s $
$t_1=2.00min\ \Rightarrow120s$
Then the distance traveled during these 120s towards the West is:

Thus, the vector of the second displacement is given by:

${\overline{r}}_2=-3000m i$

• Third displacement to the Northeast:
  

$v_3=30.0m/s $
$t_3=1.00min\ \Rightarrow60s$


Then the distance traveled during these 60s towards the Northeast, is:

Thus, the vector of this third displacement is given by:



${\overline{r}}_3=-(1272.8m)i+(1272.8m)j$

Finally the vector resulting from the displacement is given by:


Therefore, $\left|\overline{R}\right|=4865m\ \ \Rightarrow4.87Km$

The resulting vector angle is: $\theta=arctan(\frac{2327.2}{4272.8})\ \Rightarrow\ \theta=28.6°$ to the southwest

(b) The average speed


It is defined as follows:



(c) The average velocity.

It is defined as follows:
${\overline{v}}_{prom}=\frac{4.87Km}{360s}\Rightarrow\frac{4.87\times{10}^3m}{360s}$

${\overline{v}}_{prom}=13.5m/s$   along the resulting vector  $\overline{R}$