# Variable acceleration help please :)

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Thread starter 1 month ago
#1
Hi, would anyone mind helping me with this question?

Here it is: https://hosting.photobucket.com/imag...t_20.46.52.png

Here's what I've done:
a) v = 4t - t^2
0 = 4t - t^2
t = 4
integrate v to get s: s = 2t^2 - 1/3t^3 + c
c = 0
t = 4 : s = 10.7m

b) Not sure what to do for this; I've sketched a v-t graph but the roots are at 4 and 0. Can I find the area under the curve using integration and then do the same for the section between 4 and 5, then add the areas together?
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Thread starter 1 month ago
#2
(Original post by NaBrO)
b) Not sure what to do for this; I've sketched a v-t graph but the roots are at 4 and 0. Can I find the area under the curve using integration and then do the same for the section between 4 and 5, then add the areas together?
This is what I've done for (b):
https://hosting.photobucket.com/imag...-04_210723.jpg

I've got a negative area though.
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1 month ago
#3
You've got the distance from 0-4 from part a).
You make errors doing the same calculation in b).

On the return, integrate from 4 to 5 then flip the sign.
Last edited by mqb2766; 1 month ago
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1 month ago
#4
(Original post by NaBrO)
This is what I've done for (b):
https://hosting.photobucket.com/imag...-04_210723.jpg

I've got a negative area though.
You've found that the speed is instantaneously 0 at t = 4, but what happens between t = 4 and t = 5? It actually reverses its direction and moves in the other direction! So integrate from t = 0 to t= 4 to get one distance and then from 4 to 5 to get the other.
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Thread starter 1 month ago
#5
(Original post by mqb2766)
You've got the distance from 0-4 from part a).
You make errors doing the same calculation in b).

On the return, integrate from 4 to 5 then flip the sign.
May I ask where my errors are?
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1 month ago
#6
(Original post by NaBrO)
May I ask where my errors are?
The integral of 4t - t^2 is ...
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Thread starter 1 month ago
#7
(Original post by mqb2766)
The integral of 4t - t^2 is ...
2 t^2 - 1/3 t^3

Thank you!
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1 month ago
#8
(Original post by NaBrO)
2 t^2 - 1/3 t^3

Thank you!
Yeah, you did it in part a) ...
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Thread starter 1 month ago
#9
(Original post by mqb2766)
Yeah, you did it in part a) ...
I realise now...
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