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Grade 11Mechanics

Ship a is moving with velocity 30 m/s due east and Ship is with velocity 40m/s due north. Intial separation between the ships is 10 kilometre as shown in figure after what times its are closer to each other.

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Profile image of Robert
7 Years agoGrade 11
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Profile image of DISHANT
7 Years ago
 
firstly let us consider the distance of ten kilometre as D .
obviously both are moving with a constant velocity therefore after time T first body A will travel 30T metres due east  and the body B will travel 40T metres due north .
 
therefore net distance travelled by B from initial point is  (40T+D) m.
now consider a right  triangle whose vertices are having a,b at distance of 30T metres and   (40T+D) m resp.
 
using pythagoras we will get the distance at that time T tht will be distance=sqrt[ 30^2+(40T+D)^2] ….............1
 
  • using differentiation w.r.t T both sides w.r.t to T  and  making the LHS=0(concept of maxima and minima) we will be able to find the value of T for which it will be having the smallest distance .now put the obtained value of T in equation 1 to get the closest distance.