Question

Compute your average velocity in the following two cases: (A) you walk 73.2m at a speed of 1.22m/s and then run 7302m at a soeed of 3.05 m/s along a stright track. (B) you walk for 1.00min at a speed of 1.22m/s and then run for 1.00 at 3.05m/s along a stright track. (C) graph x verses t for both cases and indicate how the average velocity is found on the graph

Answer 1

a) 3.00 m/s, approximately and b) 2.135 m/s

Step-by-step explanation:

a) x1 = 73.2 m and v1 = 1.22 m/s, then

`t_1 = (x_1)/(v_1) = (73.2m)/(1.22m/s)=60 s`;

x2 = 7302 m and v2 = 3.05 m/s, then

`t_2 = (x_2)/(v_2) = (7302m)/(3.05m/s)\approx 2394.1 s`.

So the average speed is

`v_a=(x_1+x_2)/(t_1+t_2)=(73.2m+7302m)/(60s+2394.1)\approx 3m/s`.

b) t1 = 60 s and v1 = 1.22 m/s, then

`x_1 = t_1* v_1 = 60s* 1.22 m/s = 73.2 m`;

t2 = 60 s and v2 = 3.05 m/s, then

`x_2 = t_2* v_2 = 60s* 3.05 m/s = 183 m`;

So the average speed is

`v_a=(x_1+x_2)/(t_1+t_2)=(73.2m+183m)/(60s+60s)=2.135m/s`.

c) Plot cumulative distance Vs. cumulative time, the slope of the line going through the origen and the final coordinate, is the velocity