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Average acceleration and instantaneous acceleration

  • Thread starter SAGHTD
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  • #1
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Homework Statement


The acceleration of a particle moving only on a horizontal xy plane is given by a= (3t)i + (4t)j, where 'a' is in m/s*2 and t is in seconds. At t=0, the position vector r= (20m)i + (40m)j locates the particle, which then has the velocity vector v=(5m/s)i + (2m/s)j. At t=4.00s what are (1)the position vector in unit vector notation and (2) the angle between its direction of travel and the positive direction of the x-axis???


Can someone please help me i'm having a bit of problems with vector question such as this i tried a few things but it was to no use. Any help would be much appreciated :)
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi SAGHTD ! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

The standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations still apply …

you can either use them with s v and a as vectors,

or you can use them as scalars (as usual), for the x and y directions separately.

Show us what you've done, and where you're stuck, and then we'll know how to help! :smile:
 
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  • #3
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well to the get the position vector you should integrate twice
v(t)=integral a(t) dt + v(0) which is given
then integrate v to get r and r(0) is also given
 
  • #4
326
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Hi SAGHTD ! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

The standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations still apply …

you can either use them with s v and a as vectors,

or you can use them as scalars (as usual), for the x and y directions separately.

Show us what you've done, and where you're stuck, and then we'll know how to help! :smile:
uhm I know that I am only a noob here but since a is a function of t why would the constant acceleration apply?
 
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  • #5
tiny-tim
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oops!

oops! i misread it! :redface:
 
  • #6
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I'm still a bit lost on where to really start vector aren't really something i understand too well... :frown:

But still if they're asking for position vector can't i just integrate "a" twice to get it position vector?? I'm not sure if i'm really making sense but my idea seems to be a bit silly too me...
 
  • #7
tiny-tim
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Hi SAGHTD! :smile:

(just got up :zzz: …)

a is (d2x/dt2)i + (d2y/dt2)j

If you integrate once, you get (dx/dt)i + (dy/dt)j, which is v.

And if you integrate once more, you get xi + yj, which is s.

So yes, you can just integrate "a" twice to get the position vector "s".

You can either do it as a scalar, one coordinate at a time (if that feels safer :wink:), or you can do it all together, as a vector. :smile:

(ie as: a = d2s/dt2, where s = xi + yj)

Have a go! :smile:
 

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