1. At 2pm the coastguard spots a rowing dinghy 500m due south of his observation point. The dinghy has constant velocity (2i + 3j)ms^-1.
a) Find in terms of t, the position vector of the dinghy t seconds after 2pm. Answer is 2ti + j(3t-500)
b) Find the distance of the dinghy from the observation point at 2:05pm.
s = vt
s = s
v = sqrt(13) -- calculated from sqrt(2^2 + 3^2)
t = 5 x 60 = 300 -- since we want time in seconds.
therefore s = sqrt(13) x 300 = 1081.7m
however, according to the answer page, the distance is actually 721m. any ideas where i went wrong? thanks