A boat is initially at the origin,heading due east at 5ms. It then experiences a constant accelerationof(-0.2i + 0.25j)ms^-2. The unit vectors i and j are directed east and north respectively.
a)State the initial velocity of the boat as a vector.(1 mark)
b) Find an expression for the velocity of the boat t seconds after it has started to accelerate. (2 marks)
c) Find the value of t when the boat is travelling due north. (3 mark)
a) 5i ms^-1
b) v=5.2ti + 0.25tj
c) Im stuck
a)State the initial velocity of the boat as a vector.(1 mark)
b) Find an expression for the velocity of the boat t seconds after it has started to accelerate. (2 marks)
c) Find the value of t when the boat is travelling due north. (3 mark)
a) 5i ms^-1
b) v=5.2ti + 0.25tj
c) Im stuck
-
Hello
a ) the velocity vector is (5, 0) (= same as yours)
b) using v = v0 + at on the components gives
(vx, vy) = (5, 0) + (- 0.2, 0.25)*t
=(5 - 0.2t)i + (0.25t)j in your notation
your result would indicate that the velocity in + x direction increases with time. But in fact the
acceleration component in x direction is negative, so the velocity must decrease with time.
c)
the final velocity vector must be (0, vy) ( = direction north), so
(0, vy) = (5, 0) + (- 0.2, 0.25)t
which gives the equation for the x component
0 = 5 - 0.2t
0.2t = 5
t = 25 seconds
Regards
a ) the velocity vector is (5, 0) (= same as yours)
b) using v = v0 + at on the components gives
(vx, vy) = (5, 0) + (- 0.2, 0.25)*t
=(5 - 0.2t)i + (0.25t)j in your notation
your result would indicate that the velocity in + x direction increases with time. But in fact the
acceleration component in x direction is negative, so the velocity must decrease with time.
c)
the final velocity vector must be (0, vy) ( = direction north), so
(0, vy) = (5, 0) + (- 0.2, 0.25)t
which gives the equation for the x component
0 = 5 - 0.2t
0.2t = 5
t = 25 seconds
Regards