# Physics A level question help?

#1
Hi I was doing this Physics question "A girl jogs at 2.0ms^-1 in a straight line for 30 seconds, turns around and returns to her starting point 20 seconds later. What is her average velocity?"

I did average speed = d/t
So d = s*t

I got (2.0 x 30) = 60 m
And when I wanted to calculate the other time I did (2.0*20) but looking and wanted to add both of the distances together and then divide it by time, to get the speed.

However the mark scheme says you need do (2*30) + (3x20), I have no idea where the "3" came from and why we multiply by that? Can anyone please explain in depth where the three came from and why?

Thank you very much for any help
0
3 months ago
#2
(Original post by heyitsme123_12)
Hi I was doing this Physics question "A girl jogs at 2.0ms^-1 in a straight line for 30 seconds, turns around and returns to her starting point 20 seconds later. What is her average velocity?"

I did average speed = d/t
So d = s*t

I got (2.0 x 30) = 60 m
And when I wanted to calculate the other time I did (2.0*20) but looking and wanted to add both of the distances together and then divide it by time, to get the speed.

However the mark scheme says you need do (2*30) + (3x20), I have no idea where the "3" came from and why we multiply by that? Can anyone please explain in depth where the three came from and why?

Thank you very much for any help
So, you've worked out she ran 60m in one direction. It says she ran back to the starting point in 20 seconds - so what was her speed for the return 60m?
0
#3
(Original post by BlueChicken)
So, you've worked out she ran 60m in one direction. It says she ran back to the starting point in 20 seconds - so what was her speed for the return 60m?
ohhhh right. thank you! god no idea how I missed that
0
3 months ago
#4
You realize the average velocity is zero?

(Original post by heyitsme123_12)
ohhhh right. thank you! god no idea how I missed that
0
3 months ago
#5
(Original post by heyitsme123_12)
ohhhh right. thank you! god no idea how I missed that
no worries! A simple diagram and the info you know (or don't, and need to determine) written on it is good to keep things straight.

0
#6
(Original post by mqb2766)
You realize the average velocity is zero?
yes after working it out, did you manage to know it without working out? if so how did u manage to do this if u dont mind me asking
0
3 months ago
#7
(Original post by heyitsme123_12)
yes after working it out, did you manage to know it without working out? if so how did u manage to do this if u dont mind me asking
The total displacement (end-start) is zero as it returns to the initial position.
However, the average speed is non-zero which is perhaps what the working was referring to.
Last edited by mqb2766; 3 months ago
0
#8
(Original post by mqb2766)
The total displacement (end-start) is zero as it returns to the initial position.
However, the average speed is non-zero which is perhaps what the working was referring to.
Oh right!! yes that is true haha no idea how i missed that, thanks though
0
3 months ago
#9
(Original post by heyitsme123_12)
Oh right!! yes that is true haha no idea how i missed that, thanks though
Its a reasonably common one to look out for as people confuse average speed and velocity. If you return to the initial position, no matter how you do it, the average velocity is zero. So a rocket going to the moon and back (with a slight adjustment for launch/landing position) has zero average velocity.
0
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