A boat moves with a constant acceleration of(0.2i + 0.1j)ms,where i and j are directed east+north respectively

Question

Time, t is measured in seconds.

At time t= 10 the boat is at the point A, which has position vector (30i - 35j) metres.
At time t = 20, the boat is at the point B, which has a position vector (70i - 40j) metres.

a) Find the vector AB?

b)Find the velocity of the boat when t = 10?

c) Using your answer from part (b), find the velocity f the boat when t = 0?

d) Find the postion vector of the boat when t = 0?

Answer 1

a) The vector AB is just the subtraction of B - A.

B - A = (70i - 40j) - (30i - 35j) = (40i - 5j)

b) From a, we know that the boat traveled a distance of (40i - 5j) in 10 seconds. We're also given the acceleration as (0.2i + 0.1j) m/s^2.

I think the vector formula for distance says that the change in position is equal to the initial velocity times time plus 1/2 the acceleration times the time squared? (In this case, we're saying t=10 is t'=0, so t=20 becomes t'=10.)

x - x0 = v0t + 1/2at^2

40i - 5j = v0(10) + 1/2(0.2i + 0.1j)(10)^2
40i - 5j = 10v0 + 1/2(20i + 10j)
40i - 5j - (10i + 5j) = 10v0
30i - 10j = 10v0
v0 = 3i - j

So at t=10, the velocity of the boat was 3i - j m/s.

c) Now you can use the formula:

v = v0 + at

This time, we'll stick with t=0 as the reference point, so we're looking for v0. Then the velocity we just calculated was at t=10.

3i - j = v0 + (0.2i + 0.1j)(10)
3i - j - (2i + j) = v0
i - 2j = v0

d) Now we've got the "real" v0, and all we have to do is figure what the real x0 was. So we pick a point, plug in values, and solve for x0. At t=10, the position vector is 30i - 35j, so:

x = x0 + v0t + 1/2at^2
30i - 35j = x0 + (i - 2j)(10) + 1/2(0.2i + 0.1j)(10)^2
30i - 35j = x0 + 10i - 20j + 1/2(20i + 10j)
30i - 35j - (10i - 20j + 10i + 5j) = x0
x0 = 30i - 35j - 20i + 15j
x0 = 10i - 20j m

Disclaimer: please check my math. Also it's been 15 years since I had Physics...

Answer 2

a) At t = 2 velocity alongside i = 0 2 (a-2) = 0 --> a = 2 b) preliminary velocity: t = 0 --> v_0 = (4,4) acceleration: by-made from v = -2i - 4j c) x = v_0 t + acc t² = (4,4)*2 - (2,4)*4 = (0,-8) Distance is 8m.