Prob an easy question.. just missing something.

[renegade05]

Can you please help me with this problem?

A relay race is run along a straight-line track of length 300.0m running south to north. The first runner starts at the south end of the track and passes the baton to a teammate at the north end of the track. The second runner races back to the start line and passes the baton to a third runner who races 100.0 m north ward to the finish line. The magnitudes of the average velocities of the first, second, and third runners during their parts of the race are 7.30 m/s, 7.20 m/s, and 7.80 m/s, respectively. What is the average velocity of the baton for the entire race?

The answer in the book says 1.05 m/s.

But I keep getting 2.48 m/s.

Total time travelled is 40.4 seconds so wouldn’t I just use Vave = (delta x) / (delta t).

Displacement is 100 m and time is 40.4 seconds. 100/40.4 = 2.47 m/s.

I can’t figure it out.

 

[Deleted member 4993]

Can you please help me with this problem?

A relay race is run along a straight-line track of length 300.0m running south to north. The first runner starts at the south end of the track and passes the baton to a teammate at the north end of the track. The second runner races back to the start line and passes the baton to a third runner who races 100.0 m north ward to the finish line. The magnitudes of the average velocities of the first, second, and third runners during their parts of the race are 7.30 m/s, 7.20 m/s, and 7.80 m/s, respectively. What is the average velocity of the baton for the entire race?

The answer in the book says 1.05 m/s.

But I keep getting 2.48 m/s.

Total time travelled is 40.4 seconds so wouldn’t I just use Vave = (delta x) / (delta t).

Displacement is 100 m and time is 40.4 seconds. 100/40.4 = 2.47 m/s.

I can’t figure it out.

The total time is:

300/7.3 + 300/7.2 + 100/7.8 = 41.09589041 + 41.66666667 + 12.82051282 = 95.5830699 s

This should give you the correct answer.

 

[soroban]

Hello, renegade05!

A relay race is run along a straight-line track of length 300 m running south to north.
The first runner starts at the south end of the track and passes the baton to a teammate at the north end of the track.
The second runner races back to the start line and passes the baton to a third runner who races 100 m northward to the finish line.
The average velocities of the first, second, and third runners during their parts of the race are 7.30 m/s, 7.20 m/s, and 7.80 m/s, respectively.
What is the average velocity of the baton for the entire race?

The answer in the book says 1.05 m/s. . Impossible!

But I keep getting 2.48 m/s.

Total time travelled is 40.4 seconds . How did you get this?

. . so wouldn’t I just use: \(\displaystyle V_{\text{ave}} \,=\, \frac{\Delta x}{\Delta t}\)

Displacement is 100 m (No!) and time is 40.4 seconds. 100/40.4 = 2.47 m/s.

Subhotosh is absolutely correct!

I'll baby-step through the problem.


\(\displaystyle \text{The first runner ran 300 m at 7.3 m/s.}\)
. . \(\displaystyle \text{This took: }\tfrac{300}{7.3} \:\approx\;41.10\text{ seconds.}\)

\(\displaystyle \text{The second runner ran 300 m at 7.2 m/s.}\)
. . \(\displaystyle \text{This took: }\tfrac{300}{7.2} \:\approx\:41.67\text{ seconds.}\)

\(\displaystyle \text{The third runner ran 100 m at 7.8 m/s.}\)
. . \(\displaystyle \text{This took: }\tfrac{100}{7.8} \:\approx\:12.82\text{ seconds.}\)


\(\displaystyle \text{Total distance is: }300 + 300 + 100 \:=\:700\text{ m.}\)

\(\displaystyle \text{Total time is: }41.10 + 41.67 + 12.82 \,=\,95.59\text{ s.}\)

\(\displaystyle \text{Therefore, average velocity is: }\frac{700}{95.59} \;\approx\;7.32\text{ m/s.}\)


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The book's answer makes no sense . . . neither does yours.

Suppose you went on a road trip.
For a while you drove at 73 mph, later you drove at 72 mph,
and at the end you drove at 78 mph.

Is it possible that your average speed was only 10.5 or 24.8 mph?
Think about it!

 

[renegade05]

Soroban, you are incorrect! The book is correct.

The Average Velocity is 1.05m/s.

You have to remember this is displacement not distance travelled.

The baton only displaces 100 m over 95.50 s. 100/95.59 s = 1.05 m/s.

Recall
\(\displaystyle Vave = \frac{\Delta{R}}{\Delta{T}}\)

The book's answer makes no sense . . . neither does yours.

Suppose you went on a road trip.
For a while you drove at 73 mph, later you drove at 72 mph,
and at the end you drove at 78 mph.

Is it possible that your average speed was only 10.5 or 24.8 mph?
Think about it! No it would not be possible for my average speed to be that, but it would be possible for my average magnitude of my velocity to be that.