How hard is it to find a tenured position?

Find the position of the point where the boats hits the bank? Find the velocity of the boat when its position?

• The units vectors i and j are directed north and east respectively. A boat has initial boat (4i+6j)ms^-1 and an initial position of (80j+20j)m. It experiences an accleration of (-0.02i-0.04j)ms^-2 for a period of 4 minutes. After 4 minutes the boat stops accelerating and continues with a constant velocity for a further minute before hitting the bank and stopping. Find the position of the point where the boats hits the bank? Find the velocity of the boat when its position is 476i+452j.

• Answer:

  Ok this is a long one. Question 1 First of all the position where the boat hits the bank consists of two separate equation. The first equation is while the boat is accelerating and the second is while the boat is travelling at a constant velocity. This means that the position vector of the boat after it has finished accelerating must be found before the answer is present. The position vector of the point where the boat stops accelerating can be found by using s=ut+0.5at^2 s = displacement u = initial velocity a = acceleration t = time This equation is easier done as either i or j separately, not together. First the i vector. s = (4 x 240) + (0.5 x -0.02 x 240^2) = 960 + (0.5 x -0.02 x 57600) = 960 + (-0.01 x 57600) s = 960 - 576 s = 384 This is only the displacement from the initial position therefore the position vector requires +80i Therefore s = 464i Next the j vector s = (6 x 240) + (0.5 x -0.04 x 240^2) = 1440 + (-0.02 x 57600) = 1440 - 1152 s = 288j Therefore the position vector of the boat after 4 minutes in 464i + 288j At this point the final velocity must be found using v=u+at v = (4i+6j) + (-0.02i -0.04j)240 = (4+(-0.02 x 240))i + (6+(-0.04 x 240)) v = -0.8i - 3.6j Therefore by using the equation s=vt we can find the final position vector where the boat hits the bank. s = (-0.8i -3.6j) x 60 s = -48i - 216j This is only the displacement from the initial position therefore the position vector requires 264i+288j Therefore s = 216i + 72j Question 1 answered Question 2 At the point 476i+452j the displacement from the initial position = (476-80)i + (452-20)j s = 396i + 432j u = 4i+6j v = ? a = -0.02i -0.04j v^2 = u^2 + 2as This equation MUST be done separately with i and j vectors. The i vector. v^2 = 4^2 + (2 x -0.02 x 396) = 16 - 15.84 = 0.16 v = 0.4i The j vector v^2 = 6^2 +(2 x -0.04 x 432) = 1.44 v = 1.2j Therefore the velocity at point 476i+452j = (0.4i+1.2j)ms^-1 Question 2 answered