Question

A boy walks a distance of 100 meters to the right at a steady speed of 1.00 m/s. Then he stops for 30 seconds. Then returns back and walks for 2 minutes to the left at speed of 1.5m/s.
Calculate the total time of his round trip.
Calculate the total distance.
Calculate the total displacement.
Calculate the average speed.
Calculate the average velocity

Answer 1

Explanation:

$speed=\frac{Distance}{time}$

In the figure attached:

A boy travels AB distance of 100 m, with speed of 1.00 m/s

AB = 100 m

$1.00 m/s=\frac{100 m}{t_1}$

$t_1=100 s$

After reaching at point B he stops there at for 30 seconds.

$t_2= 30 s$

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=$t_3$

$1.00=\frac{100 m}{t_3}$

$t_3=100 s$

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 = ?

$t_4=2 min= 120 s$

$1.5 m/s=\frac{AC}{120 s}$

AC = 180 m

The total time of his round trip is:T

$T=t_1+t_2+t_3+t_4=100 s+30 s(stop)+ 100 s(returning)+120 s=350 s$

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:

$=\frac{AB+BA+AC}{t_1+t_2+t_3+t_4}=\frac{D}{T}$

$\frac{D}{T}=\frac{380 m}{350 s}=1.085 m/s$

The average velocity of the boy is given by:

$=\frac{AC}{t_1+t_2+t_3+t_4}=\frac{d}{T}$

$\frac{d}{T}=\frac{180 m}{350 s}=0.5142 m/s$