Question

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.

Answer 1

Answer:

a) D=-4273(x)-2327(y)

b) v=23.33m/s

c) V=-13.51m/s $28.57^o$ south of west.

Explanation:

In order to find the displacement we have to divide the motion on its X and Y components:

$d_{x1}=(25m/s)*2(min)*\frac{60s}{1min}=3000m\\d_{x2}=30m/s*1(min)*\frac{60s}{1min}*cos(45)=1273m$

The total X displacement is the algebaric sum of the movement:

$d_x=-3000m-1273m=-4273m$

We have to do the same on Y:

$d_{y1}=(20m/s)*3(min)*\frac{60s}{1min}=3600m\\d_{y2}=30m/s*1(min)*\frac{60s}{1min}*sin(45)=1273m$

$d_y=-3600m+1273m=-2327m$

D=-4273(x)-2327(y)

The angle is given by:

$\alpha =arctg(\frac{Dy}{Dx})\\\\\alpha =arctg(\frac{2327}{4273})\\\alpha=28.57^o$

That is 28.57 degrees south of west.

the total displacement magnitude is given by:

$D=\sqrt{(-4273)^2+(-2327)^2} =4865m$

The average speed is given by:

$v=\frac{(20m/s*3(min)+25m/s*2(min)+20m/s*1(min))*(60s/ min)}{6min*(60s/ min)}\\v=23.33m/s$

and the Average velocity is given by:

$|v|=\frac{4865m}{6*(60s/1min)}=13.51m/s$

The average velocity is:

$V=-13.51m/s$

with an angle of 28.57 degrees south of west.