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

Answer:

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

Explanation:

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

$t_1 = \frac{x_1}{v_1} = \frac{73.2m}{1.22m/s}=60 s$;

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

$t_2 = \frac{x_2}{v_2} = \frac{7302m}{3.05m/s}\approx 2394.1 s$.

So the average speed is

$v_a=\frac{x_1+x_2}{t_1+t_2}=\frac{73.2m+7302m}{60s+2394.1}\approx 3m/s$.

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

$x_1 = t_1\times v_1 = 60s\times 1.22 m/s = 73.2 m$;

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

$x_2 = t_2\times v_2 = 60s\times 3.05 m/s = 183 m$;

So the average speed is

$v_a=\frac{x_1+x_2}{t_1+t_2}=\frac{73.2m+183m}{60s+60s}=2.135m/s$.

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