The image the question is based on is in this link, along with the info.
http://i45.tinypic.com/zn9qv.png
What is the dinosaur's average velocity during position B and D? Unit for answer is in m/s
Please EXPLAIN to me the steps you went through in order to get the answer. I would really like to LEARN.
http://i45.tinypic.com/zn9qv.png
What is the dinosaur's average velocity during position B and D? Unit for answer is in m/s
Please EXPLAIN to me the steps you went through in order to get the answer. I would really like to LEARN.
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The purpose of this question is to test whether you know the difference between average velocity and average speed. Do you? If you cannot define and distinguish between the two you best go back and reread you assignment and this time pay particular attention to the definitions for average speed and average velocity.
Anyway the average velocity is just the final displacement dS divided by the total time spent to end up with that displacement dT. So Vavg = dS/dT.
And, here's the really important part, the total displacement is just dS = 35 m. What? After all that wandering around and the displacement is only 35 m? Yep. Dino the saur started at S0 = 5 m and ended up at S1 = 40 m. So dS = S1 - S0 = 40 - 5 = 35 m.
So now we have Vavg = dS/dT = 35/dT = ? We need dT.
And that's dT = SUM(t) where t is the time spend on each of Dino's travel segments, By eyeballing the times for each segment, I find that dT = SUM(t) = 94 seconds. (Double check I did this by eyeball.)
And there you are tadpole: Vavg = dS/dT = 35/94 = ? m/s East You can do the math. Don't forget the direction, east, as that's the nature of vectors; they have both magnitude and direction.
Anyway the average velocity is just the final displacement dS divided by the total time spent to end up with that displacement dT. So Vavg = dS/dT.
And, here's the really important part, the total displacement is just dS = 35 m. What? After all that wandering around and the displacement is only 35 m? Yep. Dino the saur started at S0 = 5 m and ended up at S1 = 40 m. So dS = S1 - S0 = 40 - 5 = 35 m.
So now we have Vavg = dS/dT = 35/dT = ? We need dT.
And that's dT = SUM(t) where t is the time spend on each of Dino's travel segments, By eyeballing the times for each segment, I find that dT = SUM(t) = 94 seconds. (Double check I did this by eyeball.)
And there you are tadpole: Vavg = dS/dT = 35/94 = ? m/s East You can do the math. Don't forget the direction, east, as that's the nature of vectors; they have both magnitude and direction.
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the answer is -.16m/s.
v = x/t
The average velocity between B and C is -1.32m/s (-45/34=1.32)
The average velocity between C and C is 1m/s (10/10=1)
To find the average of these two velocities we take the mean of the two:
(-1.32+1)/2 = -.16m/s
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v = x/t
The average velocity between B and C is -1.32m/s (-45/34=1.32)
The average velocity between C and C is 1m/s (10/10=1)
To find the average of these two velocities we take the mean of the two:
(-1.32+1)/2 = -.16m/s
.
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Point B = +25m
from there it goes to point C at -20m in 34 sec. That is 45 m in 34 sec.
Then it goes to the -10 line in 10 sec. That is 10 m in 10 sec
In total it does (45 + 10) = 55 m in (34 + 10) = 44 sec
so
55 m in 44 sec = 55/44 m/s = 5/4 m/s = 1.25 m/s
from there it goes to point C at -20m in 34 sec. That is 45 m in 34 sec.
Then it goes to the -10 line in 10 sec. That is 10 m in 10 sec
In total it does (45 + 10) = 55 m in (34 + 10) = 44 sec
so
55 m in 44 sec = 55/44 m/s = 5/4 m/s = 1.25 m/s
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Average velocity is given by this equation.
AV = (Vi + Vf)/2 . this is assuming constant acceleration
AV = (Vi + Vf)/2 . this is assuming constant acceleration