Simple question about Motion graphs/trajectory

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Two questions:

1.) When you throw a ball up and it reaches the height of its trajectory, is the velocity equal to zero only for an INSTANT or for LESS THAN A SECOND BUT MORE THAN AN INSTANT? The velocity graph is sloping straight downwards with no breaks.

2.) If you are given a x(t) graph that shows a simple parabolic motion (throw ball up and comes down) and a v(t) graph that is a straight line sloping downwards (starting above the x axis and ending below the x axis), how would you calculate the acceleration (ball is thrown on another planet)?

Any help is appreciated. Thanks.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi physics213! Welcome to PF! :smile:
1.) When you throw a ball up and it reaches the height of its trajectory, is the velocity equal to zero only for an INSTANT or for LESS THAN A SECOND BUT MORE THAN AN INSTANT? The velocity graph is sloping straight downwards with no breaks.
Only an instant (in other words, no time at all). :smile:

(Obviously, that's for a ball going straight up. If it's a parabola, then the vertical component is zero for an instant.)
2.) If you are given a x(t) graph that shows a simple parabolic motion (throw ball up and comes down) and a v(t) graph that is a straight line sloping downwards (starting above the x axis and ending below the x axis), how would you calculate the acceleration (ball is thrown on another planet)?
Hint: what is the definition of acceleration? :smile:
 
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Hi physics213! Welcome to PF! :smile:


Only an instant (in other words, no time at all). :smile:

(Obviously, that's for a ball going straight up. If it's a parabola, then the vertical component is zero for an instant.)


Hint: what is the definition of acceleration? :smile:

I know the definition of acceleration is the slope of the v(t) graph, but does that definition still hold true for a ball going up and coming down? And that acceleration would be negative right? It just doesn't seem like the acceleration would be constant all the way through.
 
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For simplicity, imagine the object is thrown upwards at a rate of exactly g. After one second, it will "decelerate" by exactly acceleration due to gravity (1sec*g=g.) -- that is, after one second it will come to a complete stop. Now what happens if we wait 1.001 seconds? Obviously it won't be EXACTLY at a stand still. Hence it only stops for an instant.
 
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tiny-tim
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I know the definition of acceleration is the slope of the v(t) graph, but does that definition still hold true for a ball going up and coming down? And that acceleration would be negative right?
It just doesn't seem like the acceleration would be constant all the way through.
When the line of the velocity graph is above the x-axis, the ball is rising.

When it is below the x-axis, the ball is falling.

But it's an unbroken line … it goes through the x-axis "without blinking"!

The acceleration, as measured on the line, is the same whether the velocity is positive or negative … which matches the constant acceleration which you know gravity has. :smile:
 

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