Explanation:
[tex]speed=\frac{Distance}{time}[/tex]
In the figure attached:
A boy travels AB distance of 100 m, with speed of 1.00 m/s
AB = 100 m
[tex]1.00 m/s=\frac{100 m}{t_1}[/tex]
[tex]t_1=100 s[/tex]
After reaching at point B he stops there at for 30 seconds.
[tex]t_2= 30 s[/tex]
After , 30 seconds he he comes back to his initial position that is A with steady speed of 1.00 m/s.
Distance covered from B to A= 100 m
Time taken by him during coming back=[tex]t_3[/tex]
[tex]1.00=\frac{100 m}{t_3}[/tex]
[tex]t_3=100 s[/tex]
After returning to to point A he turns left and moves towards point C with speed of 1.5 m/s for 2 minutes.
Distance of AC = ?
[tex]t_4=2 min= 120 s[/tex]
[tex]1.5 m/s=\frac{AC}{120 s}[/tex]
AC = 180 m
The total time of his round trip is:T
[tex]T=t_1+t_2+t_3+t_4=100 s+30 s(stop)+ 100 s(returning)+120 s=350 s[/tex]
The total distance: D
D = AB + BA (returning) + AC=100 m + 100 m + 180 m = 380 m[/tex]
The total displacement of boy:
Displacement is the shortest distance between the initial point and final point.
He first walked to point B from A and then came back to A . And after that walking to point C from A.So, the final displacement (d) is from A to C.
d = AC = 180 m
The total displacement of boy is 180 m.
The average speed of the boy is given by:
[tex]=\frac{AB+BA+AC}{t_1+t_2+t_3+t_4}=\frac{D}{T}[/tex]
[tex]\frac{D}{T}=\frac{380 m}{350 s}=1.085 m/s[/tex]
The average velocity of the boy is given by:
[tex]=\frac{AC}{t_1+t_2+t_3+t_4}=\frac{d}{T}[/tex]
[tex]\frac{d}{T}=\frac{180 m}{350 s}=0.5142 m/s[/tex]
