Answer:
a) D=-4273(x)-2327(y)
b) v=23.33m/s
c) V=-13.51m/s [tex]28.57^o[/tex] south of west.
Explanation:
In order to find the displacement we have to divide the motion on its X and Y components:
[tex]d_{x1}=(25m/s)*2(min)*\frac{60s}{1min}=3000m\\d_{x2}=30m/s*1(min)*\frac{60s}{1min}*cos(45)=1273m[/tex]
The total X displacement is the algebaric sum of the movement:
[tex]d_x=-3000m-1273m=-4273m[/tex]
We have to do the same on Y:
[tex]d_{y1}=(20m/s)*3(min)*\frac{60s}{1min}=3600m\\d_{y2}=30m/s*1(min)*\frac{60s}{1min}*sin(45)=1273m[/tex]
[tex]d_y=-3600m+1273m=-2327m[/tex]
D=-4273(x)-2327(y)
The angle is given by:
[tex]\alpha =arctg(\frac{Dy}{Dx})\\\\\alpha =arctg(\frac{2327}{4273})\\\alpha=28.57^o[/tex]
That is 28.57 degrees south of west.
the total displacement magnitude is given by:
[tex]D=\sqrt{(-4273)^2+(-2327)^2} =4865m[/tex]
The average speed is given by:
[tex]v=\frac{(20m/s*3(min)+25m/s*2(min)+20m/s*1(min))*(60s/ min)}{6min*(60s/ min)}\\v=23.33m/s[/tex]
and the Average velocity is given by:
[tex]|v|=\frac{4865m}{6*(60s/1min)}=13.51m/s[/tex]
The average velocity is:
[tex]V=-13.51m/s[/tex]
with an angle of 28.57 degrees south of west.
