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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.

User Knowdotnet
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2 Answers

3 votes

Answer:

The average velocity is defined as


v_(avg) =(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=(d)/(v)=(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=(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) =(d_(total) )/(t_(total) )=(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) =(d_(total) )/(t_(total) )=(329.90m+75.6m)/(60sec+94.8sec)\\ v_(avg) =(405.5m)/(154.8sec)= 2.62 m/s

User Toandv
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1 vote

Answer:

a)
v_(ave)=1.85(m)/(s).

b)
v_(ave)=2.61(m)/(s).

Step-by-step 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(m)/(s) \\x_2=56.8m\\v_2=3.48(m)/(s)


x=vt
t=(x)/(v)

Time walking:
t_1=45s

Time running:
t_1=16.32s

Average velocity is:
v_(ave)=(\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(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(m)/(s) \\t_2=93.6s\\v_2=3.48(m)/(s)


x=vt

Distance walking:
x_1=75.6m

Distance running:
x_2=325.72m

Average time is:
v_(ave)=(\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(m)/(s).

User MarkAWard
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