Question

Compute your average velocity in the following two cases: (a) You walk 56.8 m at a speed of 1.26 m/s and then run 56.8 m at a speed of 3.48 m/s along a straight track. (b) You walk for 1.00 min at a speed of 1.26 m/s and then run for 1.58 min at 3.48 m/s along a straight track.

Answer 1

Answer:

a) $v_{ave}=1.85\frac{m}{s}$.

b) $v_{ave}=2.61\frac{m}{s}$.

Explanation:

a) You walk 56.8 m at a speed of 1.26 m/s and then run 56.8 m at a speed of 3.48 m/s along a straight track.

We order the information:

$x_1=56.8m\\v_1=1.26\frac{m}{s} \\x_2=56.8m\\v_2=3.48\frac{m}{s}$

$x=vt$ ⇒ $t=\frac{x}{v}$

Time walking: $t_1=45s$

Time running: $t_1=16.32s$

Average velocity is: $v_{ave}=\frac{\Delta x}{\Delta t}$

For the total time elapsed:

$\Delta t=t_1+t_2=61.32\\\Delta x=x_1+x_2=113.6m$

⇒ $v_{ave}=1.85\frac{m}{s}$

b) You walk for 1.00 min at a speed of 1.26 m/s and then run for 1.58 min at 3.48 m/s along a straight track.

We order the information:

$t_1=60s\\v_1=1.26\frac{m}{s} \\t_2=93.6s\\v_2=3.48\frac{m}{s}$

$x=vt$

Distance walking: $x_1=75.6m$

Distance running: $x_2=325.72m$

Average time is: $v_{ave}=\frac{\Delta x}{\Delta t}$

For the total time elapsed:

$\Delta t=t_1+t_2=153.6s\\\Delta x=x_1+x_2=401.32m$

⇒ $v_{ave}=2.61\frac{m}{s}$.

Answer 2

Answer:

The average velocity is defined as

$v_{avg} =\frac{d_{total} }{t_{total} }$

Which means we need to find the total time and the total distance traveled.

(a)

Now, by given, we know that the person walked 56.8 meters as total, with an speed of 1.26 m/s. We can use this information to find the time that took to traveled all the way

$t=\frac{d}{v}=\frac{56.8m}{1.26 m/s}= 45.08sec$

The first part took 45.08 sec.

We repeat the process for the second part with $v=3.48m/s$

$t=\frac{56.8m}{3.48m/s}= 16.32 sec$

The second part took 16.32 seconds.

So the total distant and total time are

$d_{total}=56.8m+56.8m= 113.6m$

$t_{total}=45.08 sec+16.32sec= 61.4sec$

The average speed is

$v_{avg} =\frac{d_{total} }{t_{total} }=\frac{113.6m}{61.4sec}= 1.85m/s$

(b)

This time we have 1 minutes (which is 60 seconds) in the first part at a speed of 1.26m/s. So, the distance traveled is

$d=1.26m/s(60sec)=75.6m$

The second part took 1.58 minutes, which are (1.58x60) seconds at 3.48 m/s

$d=94.8sec(3.48m/s)=329.90 m$

So, the average velocity is

$v_{avg} =\frac{d_{total} }{t_{total} }=\frac{329.90m+75.6m}{60sec+94.8sec}\\ v_{avg} =\frac{405.5m}{154.8sec}= 2.62 m/s$