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M1 vectors q help pleasse

A particle P starts at the point with position vector Ro P moves with constant velocity vms after t seconds P is at point with position vector r

Find t if Ro = 4i+j r= 12i-11j and v= 2i-3j

a bit confused answer is t=4 but not sure how to get there if anyone could explain that would really help thankss !


also find t if Ro = -2i+3j R= 6i-3j and the speed of P is 4ms-1 :s-smilie: :s-smilie: ?!
(edited 11 years ago)
Reply 1
Avatar for lebron_23
lebron_23
2
Not sure if you're still stuck on this problem, but I'll point you (and and anyone else that stumbles onto this thread) in the right direction:

You know that the final position vector is equal to the initial postion vector plus the velocity multiplied by the time, given by r=r0+vtr = r_0 + vt so you can just sub the given values into the equation.

This leaves you with (12i−11j)=(4i+j)+(2i−3j)t(12i - 11j) = (4i +j) + (2i - 3j)t

It should be relatively straightforward, now, to work out t.

For the second question, you know that the modulus (absolute value) of a given velocity is its speed. Extending this logic, you can take the modulus of the entire equation to find the speed as working out the velocity and subbing it in is not an option. Remembering that ∣vt∣=∣v∣∗∣t∣|vt| = |v|*|t|, you can mod the equation leaving you with ∣8i−6j∣=∣vt∣|8i - 6j| = |vt|

EDIT: I may have told a fib at the end, I don't think its ∣vt∣=∣v∣∗∣t∣|vt| = |v|*|t| but rather, ∣v∣∗t|v| * t . Not too sure though, perhaps someone can check it and confirm.
(edited 11 years ago)
Reply 2
Avatar for .raiden.
.raiden.
10
Original post by lebron_23
Not sure if you're still stuck on this problem, but I'll point you (and and anyone else that stumbles onto this thread) in the right direction:

You know that the final position vector is equal to the initial postion vector plus the velocity multiplied by the time, given by r=r0+vtr = r_0 + vt so you can just sub the given values into the equation.

This leaves you with (12i−11j)=(4i+j)+(2i−3j)t(12i - 11j) = (4i +j) + (2i - 3j)t

It should be relatively straightforward, now, to work out t.

For the second question, you know that the modulus (absolute value) of a given velocity is its speed. Extending this logic, you can take the modulus of the entire equation to find the speed as working out the velocity and subbing it in is not an option. Remembering that ∣vt∣=∣v∣∗∣t∣|vt| = |v|*|t|, you can mod the equation leaving you with ∣8i−6j∣=∣vt∣|8i - 6j| = |vt|

EDIT: I may have told a fib at the end, I don't think its ∣vt∣=∣v∣∗∣t∣|vt| = |v|*|t| but rather, ∣v∣∗t|v| * t . Not too sure though, perhaps someone can check it and confirm.

After getting (12i−11j)=(4i+j)+(2i−3j)t(12i - 11j) = (4i +j) + (2i - 3j)t to find t simply make the i components equal so 12=4+2t12 = 4 + 2t this should lead to the correct answer OP:smile:

EDIT: It should work when making j components equal too
(edited 11 years ago)

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