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## Main Question or Discussion Point

I have some problem with velocity and gravitational acceleration.

Here is my question and answer as it is in my text book, which I could not understand:

A runner make one lap around a 200m track in a time of 25s. What were the runner's (a) average speed and (b) average velocity?

(a)From the definition,

Average speed=[tex]\frac{distance traveled}{time taken}[/tex]

Average speed=[tex]\frac{200m}{25s}[/tex]=8.0m/s

(b)Because the run ended at the starting point, the displacement vector from starting point to end point has zero length. Since [tex]\overline{v}[/tex]=[tex]\frac{\overline{s}}{t}[/tex],

[tex]\overline{v}[/tex]=[tex]\frac{0m}{25s}[/tex]=0m/s

Now, here is my that question in which I have problem:

A ball that is thrown vertically upward on the Moon returns to its starting point in 4.0s. The acceleration due to gravity there is 1.60m/s

Let us take up as positive. For the trip from beginning to end, y=0(it ends at the same level it started at), a=-1.60m/s

0=v

from which v

As you see in 1st question that when calculating velocity then we have distance is 0 while in the 2nd question it is clearly said to calculate ball's original speed.

If we are calculating ball's original speed then why we took y=0? While we are calculating speed in which direction doesn't matter.

Can anyone explain this for me???

Here is my question and answer as it is in my text book, which I could not understand:

A runner make one lap around a 200m track in a time of 25s. What were the runner's (a) average speed and (b) average velocity?

(a)From the definition,

Average speed=[tex]\frac{distance traveled}{time taken}[/tex]

Average speed=[tex]\frac{200m}{25s}[/tex]=8.0m/s

(b)Because the run ended at the starting point, the displacement vector from starting point to end point has zero length. Since [tex]\overline{v}[/tex]=[tex]\frac{\overline{s}}{t}[/tex],

[tex]\overline{v}[/tex]=[tex]\frac{0m}{25s}[/tex]=0m/s

Now, here is my that question in which I have problem:

A ball that is thrown vertically upward on the Moon returns to its starting point in 4.0s. The acceleration due to gravity there is 1.60m/s

^{2}downward. Find the ball's original speed.Let us take up as positive. For the trip from beginning to end, y=0(it ends at the same level it started at), a=-1.60m/s

^{2}, t=4.0s. We use y=v_{iy}t+½at^{2}to find0=v

_{iy}(4.0s)+½(-1.60m/s^{2})(4.0s)^{2}from which v

_{iy}=3.2m/s.As you see in 1st question that when calculating velocity then we have distance is 0 while in the 2nd question it is clearly said to calculate ball's original speed.

If we are calculating ball's original speed then why we took y=0? While we are calculating speed in which direction doesn't matter.

Can anyone explain this for me???

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