h26
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just part iii so I am sorta there - I got a is -0.6i +j but was wondering can you get a in scalor form for this q -ms seems to suggest otherwise. also that change in velocity bit too is confusing me.
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a is a vector, so you can't get it in scalar form.
using the basic suvat, v=u+at, you can get the velocity from this by just multiplying the acceleration vector by 10.
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h26
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(Original post by baznoy)
a is a vector, so you can't get it in scalar form.
using the basic suvat, v=u+at, you can get the velocity from this by just multiplying the acceleration vector by 10.
but for a you can just do sqrt( 3^2 +5^2) all over 5 to get a = 1.17 -->won't that work? An how do you know it is a vector?
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(Original post by h26)
just part iii so I am sorta there - I got a is -0.6i +j but was wondering can you get a in scalor form for this q -ms seems to suggest otherwise. also that change in velocity bit too is confusing me.
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The forces are given as vectors, so the acceleration you get from Newton's 2nd Law will be a vector, i.e. it has a magnitude (in ms^-2) which you can find in the same way as you find the magnitude of any vector (i.e. square root of the sum of the squares of the components), but it also has a direction.

Acceleration is by definition the change in velocity per unit of time.
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(Original post by Prasiortle)
The forces are given as vectors, so the acceleration you get from Newton's 2nd Law will be a vector, i.e. it has a magnitude (in ms^-2) which you can find in the same way as you find the magnitude of any vector (i.e. square root of the sum of the squares of the components), but it also has a direction.

Acceleration is by definition the change in velocity Dr unit of time.
Ah ok thanks so much and how do you work out the change in velocity bit. Sorry my computer is beimg extra slow today
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(Original post by h26)
Ah ok thanks so much and how do you work out the change in velocity bit. Sorry my computer is beimg extra slow today
Acceleration gives the change in velocity per second, so to get the change in velocity for 10 seconds, you simply multiply the acceleration by 10.
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(Original post by Prasiortle)
Acceleration gives the change in velocity per second, so to get the change in velocity for 10 seconds, you simply multiply the acceleration by 10.
oh so you dont use any suvat equations?
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(Original post by h26)
oh so you dont use any suvat equations?
You could use v=u+at with u=0, a=the acceleration, and t=10, but you don't need to - it's much simpler to just use the definition of acceleration. The fact that you wanted to use a SUVAT equation seems to suggest that you might be one of those students who prefers to just rote-learn methods and formulae and blindly plug in the numbers, rather than actually understanding what the concepts mean.
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(Original post by Prasiortle)
You could use v=u+at with u=0, a=the acceleration, and t=10, but you don't need to - it's much simpler to just use the definition of acceleration. The fact that you wanted to use a SUVAT equation seems to suggest that you might be one of those students who prefers to just rote-learn methods and formula and blindly plug in the numbers, rather than actually understanding what the concepts mean.
so sorry for the misunderstanding - I actually try to learn as many ways as possible to do a question so I understand it fully
With simple and consise method you advised:
acceleration is teh change in velocity in 1 second, so to find the change in velocity after 10 seconds you just multiply the acceleration by 10seconds so i understand it

However for the suvat way how do you know u = 0 -this is what confused me
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(Original post by h26)
so sorry for the misunderstanding - I actually try to learn as many ways as possible to do a question so I understand it fully
With simple and consise method you advised:
acceleration is teh change in velocity in 1 second, so to find the change in velocity after 10 seconds you just multiply the acceleration by 10seconds so i understand it

However for the suvat way how do you know u = 0 -this is what confused me
We're interested in the change in velocity, so it doesn't matter where you start - all we care about is the difference between the starting velocity and ending velocity. Thus you could say in general v = u + at and what we care about is the difference v - u, so we rearrange to get v - u = at, and thus whatever value you take as u (e.g. you could take your starting point as the time when the particle has already accelerated up to 10 m/s, if you wanted to), the value of v will be greater than it by a*t.
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(Original post by Prasiortle)
We're interested in the change in velocity, so it doesn't matter where you start - all we care about is the difference between the starting velocity and ending velocity. Thus you could say in general v = u + at and what we care about is the difference v - u, so we rearrange to get v - u = at, and thus whatever value you take as u (e.g. you could take your starting point as the time when the particle has already accelerated up to 10 m/s, if you wanted to), the value of v will be greater than it by a*t.
this way is rather confusing
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(Original post by h26)
this way is rather confusing
Do you understand that
1. We want the change in velocity, i.e. the end velocity minus the start velocity, i.e. v-u
and 2. The equation v = u + at can be rearranged to v - u = at.

Thus it doesn't matter what u you choose, v will always come out greater than u by a*t. In this case, we set t = 10 to get the answer.
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(Original post by Prasiortle)
Do you understand that
1. We want the change in velocity, i.e. the end velocity minus the start velocity, i.e. v-u
and 2. The equation v = u + at can be rearranged to v - u = at.

Thus it doesn't matter what u you choose, v will always come out greater than u by a*t. In this case, we set t = 10 to get the answer.
ohhhhhhhhhhh i seee thanks
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