# Kinematics :(

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#1
A particle travels in a straight line and decelerates uniformly at 2m/s^2. When t=0 its velocity is u m/s and when t=100 its velocity is -v m/s (where u &amp;amp;gt; v &amp;amp;gt; 0). The average speed of the particle over the 100 seconds is 62.5 m/s. Find the values of u and v.

Okay so this is what I've done so far.

S: 6250
U: u
V: -v
A: -2
T: 100

D/100= 62.5 therefore D=6250

I did simple suvat but got the wrong answer so I'm assuming thats not the right method.

Any help would be much appreciated!
0
2 years ago
#2
(Original post by leverarch)
A particle travels in a straight line and decelerates uniformly at 2m/s^2. When t=0 its velocity is u m/s and when t=100 its velocity is -v m/s (where u > v > 0). The average speed of the particle over the 100 seconds is 62.5 m/s. Find the values of u and v.

Okay so this is what I've done so far.

S
U
V: u
A: -2
T: 0

S: 6250
U
V: -v
A: -2
T: 100

D/100= 62.5 therefore d=6250

I did simple suvat for the second but got the wrong answer so I'm assuming thats not the right method.

Any help would be much appreciated!
It'd be better if you worked out the distance it travelled in the 100 seconds span, and then divided that value by 100.

Where di you even get d and D?

Try to model the movement graphically on a graph
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#3
(Original post by RickHendricks)
It'd be better if you worked out the distance it travelled in the 100 seconds span, and then divided that value by 100.

Where di you even get d and D?

Try to model the movement graphically on a graph
Sorry d and D are supposed to the same. I did distance/time=speed. I graphed it but still can’t figure it out
0
2 years ago
#4
(Original post by leverarch)
Sorry d and D are supposed to the same. I did distance/time=speed. I graphed it but still can’t figure it out
how does your graph look like?

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#5
(Original post by RickHendricks)
how does your graph look like?

https://goo.gl/images/Dsq19b Graph looks like this.

U= 150
V= 50
0
2 years ago
#6
(Original post by leverarch)
https://goo.gl/images/Dsq19b Graph looks like this.

U= 150
V= 50
0
#7
(Original post by RickHendricks)
Literally just subbing in values in s=ut+1/2at^2 and s=vt-1/2at^2
0
2 years ago
#8
(Original post by leverarch)
Literally just subbing in values in s=ut+1/2at^2 and s=vt-1/2at^2
what were the values of U and V that you got??

0
#9
(Original post by RickHendricks)
what were the values of U and V that you got??

For U I got 162.50, but the actual answer is 150. And I got the same answer for V which clearly can't be right, correct answer is 50.

Do you know how to do it?
0
#10
BUMP

can anyone else help?
0
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