The velocity of a ball
Hello there
I wonder if anyone may be able to advise me?
I've been working through the following problems:
I'm unsure whether I understand this concept and therefore whether I've answered these at all correctly?
i) >7.5 and <2.5?
ii) >2 and < 7.5?
iii) At 0 it has a positive velocity, so moves forwards to 2.5?
iv) At 5, the ball has a negative velocity, so moves backwards towards 2.5?
v) at 10, it has a positive velocity so moves forwards to the next fixed point?
This is a rough drawing I made for vi. I know the scaling and steepness of the slopes isn't precise. But am I on the right track?
I'm not sure that I understand. If anyone might be able to offer me some feedback, I'd be very appreciative.
Thank you
I wonder if anyone may be able to advise me?
I've been working through the following problems:
I'm unsure whether I understand this concept and therefore whether I've answered these at all correctly?
i) >7.5 and <2.5?
ii) >2 and < 7.5?
iii) At 0 it has a positive velocity, so moves forwards to 2.5?
iv) At 5, the ball has a negative velocity, so moves backwards towards 2.5?
v) at 10, it has a positive velocity so moves forwards to the next fixed point?
This is a rough drawing I made for vi. I know the scaling and steepness of the slopes isn't precise. But am I on the right track?
I'm not sure that I understand. If anyone might be able to offer me some feedback, I'd be very appreciative.
Thank you
(edited 9 years ago)
Those answers all look correct, except the last one. The main point with question v) is that every time the velocity changes direction, this represents a peak/trough. Since x=0 starts with positive velocity it must be downhill, x=2.5 must be a trough as velocity=0, 2.5<x<7.5 must be uphill as the velocity is negative, x=7.5 must be a peak and x>7.5 must be downhill again. So if you shift your graph to the left by 2.5 then it will be correct.
Hope this helps!
Hope this helps!
Original post by Ggdf
i) >7.5 and <2.5?
ii) >2 and < 7.5?
i) >7.5 and <2.5?
ii) >2 and < 7.5?
i) x cannot be > 7.5 and <2.5, you mean "or" there, i.e. >7.5 or <2.5?
ii) I presume that is a typo, and you meant >2.5 and < 7.5, and you do mean "and" there as x must be in the interval (2.5,7.5)
vi) If the velocity is constant, i.e. the first section, what does that tell you about the slope, or lack of?
Original post by tomtjl
Those answers all look correct, except the last one. The main point with question v) is that every time the velocity changes direction, this represents a peak/trough. Since x=0 starts with positive velocity it must be downhill, x=2.5 must be a trough as velocity=0, 2.5<x<7.5 must be uphill as the velocity is negative, x=7.5 must be a peak and x>7.5 must be downhill again. So if you shift your graph to the left by 2.5 then it will be correct.
Hope this helps!
Hope this helps!
Thank you very much for your reply.
Does this look better?
Is the area above 0 on the y axis positive (<2.5 and >7.5)? And the area below 0 >2.5 and <7.5 negative?
Original post by Ggdf
Thank you very much for your reply.
Does this look better?
Is the area above 0 on the y axis positive (<2.5 and >7.5)? And the area below 0 >2.5 and <7.5 negative?
Does this look better?
Is the area above 0 on the y axis positive (<2.5 and >7.5)? And the area below 0 >2.5 and <7.5 negative?
That looks perfect, and yes the area is negative if it is below the x axis
.Original post by tomtjl
That looks perfect, and yes the area is negative if it is below the x axis .
.Thank you very much, I really appreciate your help!
I was a little confused by ghostwalker's post and was unsure if I had made a mistake.Quick Reply
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