Find the time in which distance between ships is minimum.
Answer with explainations please.
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let's say the distance between the two ships will be minimum after 't' time...
so...
minimum distance(S) will be...
S= √[{100-(10t)}²+{10t}²]
S= √[10000+100t²-2000t+100t²]
S= √[200t²-2000t+10000]
S= √[200(t²-10t+50)]
so for minimum value of S
(dS/dt)=0
which will let us to ,
100-10t=10t
t=5seconds
so...
minimum distance(S) will be...
S= √[{100-(10t)}²+{10t}²]
S= √[10000+100t²-2000t+100t²]
S= √[200t²-2000t+10000]
S= √[200(t²-10t+50)]
so for minimum value of S
(dS/dt)=0
which will let us to ,
100-10t=10t
t=5seconds
aabhamisra:
Sorry but the velocity is in two different directions so how we can use the vale of initial velocity as 10 and also its not a case of free fall so how can we assume g as 10 m/s^2
what?!
initial velocity is given 10kmph
and I haven't assumed any value of g in it...
See one ship is moving along x axis the other one along y axis.
so what?
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