Question

A radio controlled model boat is placed in a lake at the point O with initial velocity 0.6j ms^-1 and its velocity after 15 seconds was measured as (10.5i -0.9j) ms^-1. The acceleration of the boat is modelled as being constant.

Find an expression for the position of the boat at time t. For what value of t was the boat north-East of O. For what value of t was the boat travelling in the direction north-east?

The boat lost contact with the radio transmitter when it was 100m from O. Assuming that acceleration remained constant, verify that contact was lost before 17 seconds had elapsed.

Answer 1

a) Find an expression for the position of the boat at time t

xo = 0, yo = 0

vox = 0

voy = 0.6 m/s

from the given data

ax = (vx - vox)/t = (10.5 - 0)/15 = 0.70 m/s^2

ay = (vy - voy)/t = (-0.9 - 0.6)/15 = -0.1 m/s^2

the position of boat at time t,

x = xo +vox*t + (1/2)*ax*t^2

= 0 + 0 + (1/2)*0.7*t^2

= 0.35*t^2

y = yo + voy*t + (1/2)*ay*t^2

= 0 + 0.6*t + (1/2)*(-0.1)*t^2

= 0.6*t - 0.05*t^2

b) For what value of t was the boat north-East of O

when x = y the boat will be north-east

x = y

0.35*t^2 = 0.6*t - 0.05*t^2

==> t = 1.5 s

c) For what value of t was the boat travelling in the direction north-east?

when vx = vy , the boat will be travelling towards north-east

vx = vy

vox + ax*t + voy + ay*t

0 + 0.7*t = 0.6 + (-0.1)*t

==> t = 0.75  

d) at t = 17s,

x = 0.35*t^2

= 0.35*17^2

= 101.15 m

y = 0.6*t - 0.05*t^2

= 0.6*17 - 0.05*17^2

= -4.25 m

d = sqrt(x^2 + y^2)

= sqrt(101.15^2 + 4.25^2)

= 101.2 m > 100 m

so, the contact was lost before 17 seconds had elapsed.