Question

Ship A is 10 km due West of ship B. Ship A is heading directly North at a speed of 30 km/h while ship B is heading in a direction 60° west of north at a speed of 20 km/h.a. Determine the magnitude of the velocity of ship B related to ship A.b. What will be their distance of closest approach?

Answer 1

Solution :-

Given ,

Va= 30km

Vb = 20 km

relative velocity of B and A =

Vba = √Vb - VA

Vba = √Vb^2 + Va^2 - 2VbVaCOS60°

suitable the values and solve

we get Vba = 10√7 km/ h

let the distance between the two ship be D at time t.

D = [( Vat - Vbt COS60°)^2 + ( 10 Vbt sin 60° )^2 ]^ 1/2 ---------------(I)

the distance of closest approach can be found by setting dD/dt = 0

t = √3/7h

on substitute the above of t in equation (I) we get

Dmin = 20/√7 = 7.56 km

=====================
@GauravSaxena01

Answer 2

For any arbitrary motion in space, which of the following relations are true: a. $V_{average}= (\frac{1}{2})[v(t_1)+v(t_2)]$ b. $V_{average}=\frac {[r(t_2)-r(t_1)]} {(t_2-t_1)}$ c. v (t) = v (0) + at d. r (t) = r (0) + v (0) t + (1/2) at² e. $a_{average}=\frac {[v(t_2)-v(t_1)]} {(t_2-t_1)}$ (The 'average' stands for average of the quantity over the time interval $t_1 to t_2$)