A boat moves with constant acceleration, so that at time t its velocity is given by v = 2(a - t)i + 4(3 - t)j?

Question

where a is a constant and i and j are unit vectors directed north and east respectively.

a) The boat is heading due north when t = 2. Find a?

b) Find:
i) the initial velocity of the boat?
ii) the acceleration of the boat?

c) Find the distance between the initial position of the boat and its position when t = 2??

Answer 1

Firstly, your definition of i and j goes against convention, and I don't think the question is possible as it is. It should read "east and north respectively"

a)North=>no i component of velocity at t=2
=>2(a-2)=0
2a-4=0
a=2

b)i) initial velocity is velocity at t=0
v=2(2-0)i + 4(3-0)j
=4i+12j
(or the square root of (4sqrd+12sqrd)

b)ii)To find the acceleration of the boat you could use several techniques, I will look at i and j components of acceleration separately, using velocities already established for t=0 and t=2
v=u+at vi=ui+ait vj=uj+ajt THIS "a" (ACCELERATION) IS A DIFFERENT "a" FROM THE CONSTANT USED ABOVE
i component:0=4+a(2) => ai=-2
j component:4(3-2)=12+aj(2) => 4=12+2aj => aj=-4
So the acceleration is -2i-4j or the square root of (2sqrd+4sqrd)

c)s=ut+1/2at^2
Still working with each component separately........
si=uit+1/2ait^2, sj=ujt+1/2at^2
si=4(2)+1/2(-2)(2^2)
=8-4=4
si=4

sj=12(2)+1/2(-4)(2^2)
=24-8
sj=12

So the distance travelled is 4i+12j
or the square root of (4sqrd+12sqrd)

Answer 2

How can it be due north with t=2, when its east component 4(3-2)=4>0?

Answer 3

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Answer 4

a) At t = 2 speed along i = 0
2 (a-2) = 0 --> a = 2

b) initial velocity: t = 0 --> v_0 = (4,4)
acceleration: derivative of v = -2i - 4j

c) x = v_0 t + acc t² = (4,4)*2 - (2,4)*4 = (0,-8)
Distance is 8m.

Answer 5

This is not a bloody quiz site.
I just open the throttles, and it goes quicker.
What more do you need ?
Bob.