In need of M1 help. Edexcel mechanics (May 2003 paper)?

Question

Hola, basically I'm doing M1 and I kind of hate maths at the moment so bare with me if I ask any ridiculous really obvious questions.

I've got an old exam question from 2003 & our teacher gave us all the final answer for it but I'm confused as to how she got it.

5. A particle P moved with constant acceleration (2i-3j)m/s^-2. At time t seconds, it's velocity is v/m/s^-1. When t=0, v=-2i+7j.

a. Find the value of t when P is moving parallel to the vector i. (I've done this bit & got it correct. The answer is t= 2 and a third seconds.)

b. Find the speed of P when t=3. (this is the bit I'm ridiculously stuck on. I don't understand how to get to the answer, is there a formula or something I should know about that and I'm just being a bit dense?)

c. Find the angle between the vector j and the direction of motion of P when t=3. (Haven't yet tried this, but I suck at maths so all help would be appreciated.)

Thank you for your help in advance, even if you have to roll your eyes at my stupidity please help me:)

gobob3000 · Accepted Answer

a = 2i - 3j
v(0) = 2i + 7j

v(t) = v(0) + at = 2i + 7j + 2ti - 3tj = (2 + 2t)i + (7 - 3t)j

a) Parallel to i means j = 0

so 7 -3t = 0 or t = 2 1/3

b) Speed when t = 3

v =  (2 + 2t)i + (7 - 3t)j = (2 + 2*3)i + (7 - 3*3)j = 8i - 2j

Speed (P) is simply how fast it is going, independent of direction

i and j are at right angles to each other like the x axis  and y axis, so use the Pythagorean theorem...

P^2 = 8^2 + (-2)^2 = 64+4 = 68

P = sqrt(68)

c) Let i be along the normal "x" axis and j be along the normal "y" axis

Draw a triangle from the origin with one side 8 units in the i (x) direction and then the "opposite" side -2 units in the j (y) direction. The hypotenuse is, as we saw sqrt(68).

The angle is arctan(-2/8) or about -14 degrees

[If "i" is really the "y" axis and "j" is the "x"axis  you'll just need to flip the drawing around.]

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In need of M1 help. Edexcel mechanics (May 2003 paper)?

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