# Thread: Mechanics: Vectors Help Needed

1. ## Mechanics: Vectors Help Needed

Question:
Two Ships:

Ship P: t=0; r= (20i + 30j) km, t=3; r= (29i + 34j) km
Ship Q: t=0; r= (14i - 6j) km and v=12j kmh^-1

at time t hours after midnight, position vectors of P and Q are p and q km respectively

a) Find velocity of P in terms of i and j
b) Find expression for p and q in terms of t, i and j
c) at time t hours after midnight, distance between P and Q is d km
Find expression for PQ to show that

$\displaystyle d^2=25t^2-92t +292$

a) Velocity of P is 3i +8j kmh^-1
b) p = (20 + 3t)i + (10+ 8t)j
q = 14i + (12t-6)j
c) PQ = (3t - 6)i + (4t -16)j
Therefore $\displaystyle d^2=(3t-6)^2 + (4t-16)^2$

However, This = $\displaystyle 25t^2 - 164t + 292$

Can anyone see where i have gone wrong, or if the question was wrong to start with?

2. ## Error

Your error is in the velocity of P; this propagated into the rest. The position vector of P is $\displaystyle (20\mathbf{\hat{i}} + 30\mathbf{\hat{j}})$ at t=0, and is $\displaystyle (29\mathbf{\hat{i}} + 34\mathbf{\hat{j}})$ at t=3. You had correctly that the x component of the velocity is $\displaystyle v_x=\frac{\Delta{x}}{\Delta{t}}=\frac{29-20}{3-0}=3$, but your y component is wrong: $\displaystyle v_y=\frac{\Delta{y}}{\Delta{t}}=\frac{34-30}{3-0}=\frac{4}{3}$

So the velocity of P is $\displaystyle (3\mathbf{\hat{i}} + \frac{4}{3}\mathbf{\hat{j}})$ km/h,
and the rest follows from there.

I hope this helped.

--Kevin C.