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Mechanics and Vector Help

I'm trying to do some revision before classes start again tomorrow but I'm struggling a little bit with this one, any help would be greatly appreciated! I've included my working so far:

A particle moves in 2D with a vector velocity which at time t is ' v = (-0.4ti + 5j) m/s '.

At time t=0 the particle is at (3,2)

a) Find the acceleration vector at time t

b) Find the position vector at time t

c) Find the speed at time t = 10

I found the acceleration vector at time t to be -0.4i m/s^2

I can't figure out how to find the position vector at time t and also not sure how I'd carry on to do part c.

I've integrated to get: 'r = -0.2t²i + 5tj + c' and i know at time t=0, 'r = 3i + 2j' but I'm not sure what to do to finish off the question. Thank you for any help! I know I have to substitute it in to find c, but I think I'm being a little silly and I'm not sure how to sub in and solve, when t is still in the first equation :confused:

Reply 1

ghostwalker

Original post by chris35

I've integrated to get: 'r = -0.2t²i + 5tj + c' and i know at time t=0, 'r = 3i + 2j' but I'm not sure what to do to finish off the question. Thank you for any help! I know I have to substitute it in to find c, but I think I'm being a little silly and I'm not sure how to sub in and solve, when t is still in the first equation

So, at t=0

r= 3i + 2j =-0.2x0²i + 5x0j + c = 0+c = c
And so r= -0.2t²i + 5tj +3i + 2j

And just group the i's and j's for good measure.

(edited 11 years ago)

Reply 2

chris35

Okay so I'd end up with 'r = (3-0.2t²)i + (5t+2)j'

I was wondering for part c, I assume I just sub in 10 to the equation for v, but how do I know the values of i and j at t=10?

Reply 3

ghostwalker

Original post by chris35

Okay so I'd end up with 'r = (3-0.2t²)i + (5t+2)j'

I was wondering for part c, I assume I just sub in 10 to the equation for v, but how do I know the values of i and j at t=10?

i,j are perpendicular unit vectors representing 1 m/s in the appropriate direction.

This is true for all value of t.

(edited 11 years ago)

Reply 4

chris35

Oh, so I just sub in t = 10 to get v = 1 m/s? Seems very easy!

Reply 5

ghostwalker

Original post by chris35

Oh, so I just sub in t = 10 to get v = 1 m/s? Seems very easy!

Too easy.

What have you done to get 1 m/s?

Edit: Don't forget i and j represent vectors at right angles to each other, so you'd need to use Pythagoras to work out the magnitude of the combined velocities

(edited 11 years ago)

Reply 6

chris35

Oh I assumed you meant i and j were just going to be 1, so I subbed 10 into 'v = (-0.4ti + 5j)' with i and j as 1, to get -4 + 5 = v, so v = 1. I'm guessing thats wrong though...

Reply 7

ghostwalker

Original post by chris35

Oh I assumed you meant i and j were just going to be 1, so I subbed 10 into 'v = (-0.4ti + 5j)' with i and j as 1, to get -4 + 5 = v, so v = 1. I'm guessing thats wrong though...

They represent 1 m/s in the appropriate direction.

Think of the -4i and the 5j as two sides of a right-angled triangle, and you need to find the length of the hypotenuse.