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The horizontal unit vectors i and j are in the directions east and north respectively.
A model ship of mass 2kg is moving so that its acceleration vector ams^-2 at time t seconds is given by
a=3(2t-5)i+4j. When t=T, the magnitude of the horizontal force acting on the ship is 10N.
Find the possible values of T.
If someone could tell how to work through it that would be great because i have no idea where to start.
A model ship of mass 2kg is moving so that its acceleration vector ams^-2 at time t seconds is given by
a=3(2t-5)i+4j. When t=T, the magnitude of the horizontal force acting on the ship is 10N.
Find the possible values of T.
If someone could tell how to work through it that would be great because i have no idea where to start.
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#2
(Original post by emma31x12)
The horizontal unit vectors i and j are in the directions east and north respectively.
A model ship of mass 2kg is moving so that its acceleration vector ams^-2 at time t seconds is given by
a=3(2t-5)i+4j. When t=T, the magnitude of the horizontal force acting on the ship is 10N.
Find the possible values of T.
If someone could tell how to work through it that would be great because i have no idea where to start.
The horizontal unit vectors i and j are in the directions east and north respectively.
A model ship of mass 2kg is moving so that its acceleration vector ams^-2 at time t seconds is given by
a=3(2t-5)i+4j. When t=T, the magnitude of the horizontal force acting on the ship is 10N.
Find the possible values of T.
If someone could tell how to work through it that would be great because i have no idea where to start.
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(Original post by Muttley79)
We cab't give you a solution as that is against the rules. Try using F = ma when t = T and post what you get.
We cab't give you a solution as that is against the rules. Try using F = ma when t = T and post what you get.
10= 2x(3(2t-5)i+4j)
10=12Ti-30i+8j
Am I completely off?😬
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(Original post by profkerubo)
i can help you via email
i can help you via email
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#6
(Original post by emma31x12)
Oh i tried this before and i thought i was wrong: so i did
10= 2x(3(2t-5)i+4j)
10=12Ti-30i+8j
Am I completely off?😬
Oh i tried this before and i thought i was wrong: so i did
10= 2x(3(2t-5)i+4j)
10=12Ti-30i+8j
Am I completely off?😬
Equate to 10 and solve for T
Last edited by mqb2766; 1 month ago
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#7
(Original post by profkerubo)
i can help you via email
i can help you via email
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#8
(Original post by emma31x12)
Thank you- I sent you a message w my email
Thank you- I sent you a message w my email
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(Original post by mqb2766)
use Pythagoras to find the magnitude of the corresponding force?
Equate to 10 and solve for T
use Pythagoras to find the magnitude of the corresponding force?
Equate to 10 and solve for T
Last edited by emma31x12; 1 month ago
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#10
Newton's Second Law alongside some vector work (i and j)
Last edited by TheBrightSavage; 1 month ago
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#11
(Original post by TheBrightSavage)
The Force given is the horizontal force not the total force so I think you solve it like this:
....
The Force given is the horizontal force not the total force so I think you solve it like this:
....
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#12
(Original post by TheBrightSavage)
Could I PM him/her the solution?
Could I PM him/her the solution?
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#13
(Original post by TheBrightSavage)
The Force given is the horizontal force not the total force so I think you solve it like this:
So you know the horizona ...
The Force given is the horizontal force not the total force so I think you solve it like this:
So you know the horizona ...
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#14
(Original post by emma31x12)
Sorry, what do you mean by corresponding force? Do you mean that i should separate the vector given for acceleration into i and j?
Sorry, what do you mean by corresponding force? Do you mean that i should separate the vector given for acceleration into i and j?
Btw - if you change a response, it's better to do a new reply. Editing a previous one doesn't produce a notification.
Last edited by mqb2766; 1 month ago
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(Original post by mqb2766)
No. Get the magnitude of the acceleration (pythagoas) and multiply by 2 to get the (scalar) force, which is 10. It's easier to keep things squared up. Note that in your original reply, you set a scalar (10) equal to a vector (i+j) which doesn't really make sense.
Btw - if you change a response, it's better to do a new reply. Editing a previous one doesn't produce a notification.
No. Get the magnitude of the acceleration (pythagoas) and multiply by 2 to get the (scalar) force, which is 10. It's easier to keep things squared up. Note that in your original reply, you set a scalar (10) equal to a vector (i+j) which doesn't really make sense.
Btw - if you change a response, it's better to do a new reply. Editing a previous one doesn't produce a notification.
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(Original post by mqb2766)
No. Get the magnitude of the acceleration (pythagoas) and multiply by 2 to get the (scalar) force, which is 10. It's easier to keep things squared up. Note that in your original reply, you set a scalar (10) equal to a vector (i+j) which doesn't really make sense.
Btw - if you change a response, it's better to do a new reply. Editing a previous one doesn't produce a notification.
No. Get the magnitude of the acceleration (pythagoas) and multiply by 2 to get the (scalar) force, which is 10. It's easier to keep things squared up. Note that in your original reply, you set a scalar (10) equal to a vector (i+j) which doesn't really make sense.
Btw - if you change a response, it's better to do a new reply. Editing a previous one doesn't produce a notification.
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