Average acceleration and instantaneous acceleration

[SAGHTD]

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 :)

 

[tiny-tim]

Welcome to PF!

Hi SAGHTD ! Welcome to PF!

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

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! ​

 

[madah12]

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

 

[madah12]

Hi SAGHTD ! Welcome to PF!

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

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! ​

uhm I know that I am only a noob here but since a is a function of t why would the constant acceleration apply?

 

[tiny-tim]

oops!

oops! i misread it!

 

[SAGHTD]

I'm still a bit lost on where to really start vector aren't really something i understand too well...

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...

 

[tiny-tim]

Hi SAGHTD!

(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 ), or you can do it all together, as a vector.

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

Have a go! ​