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Lcoohill
Two particles A and B have velocities 3i and vj respectively (in m/s). (a) Find rB|A, the position of B relative to A for all t given that rB|A(t = 0) = −9i+6j (in meters). (b) Find the value of v such that A and B collide. (c) If v = 1 m/s, ﬁnd the time and distance when A and B are closest together.
first, you need to find \(r_B(t)\) and \(r_A(t)\) through integration and then, since they are vectors,|dw:1354622967278:dw| \(r_B(t)-r_A(t)\) is the relative distance. You follow?
yes i get that alright i know you integrate -9i+6j to -9ti+6tj ?? not sure exactly . do i integrate A and B ? bit confused with the vj ??
vb - va = 0i - v-vj ??
you integrate \(v_A(t)=3i\) to get \(r_A(t)=3ti+C_A\) and \(v_B(t)=vj\) to get \(r_B(t)=vtj+C_B\) \(r_B(t)-r_A(t)=vtj+C_B-3ti-C_A=vtj-3ti+C\)
\(r_B(0)−r_A(0)=C=−9i+6j\)
so, \(r_B(t)−r_A(t)=-3(t+3) i + (vt+6) j\) :) you follow? you'll have to do them one by one
kind of sorry was thought different way in college and am just really confused i get what u are saying to integrate both to get position and then take a from b
ah. perhaps my solution was a little...um, confusing. :) so you've got rB|A? for the second question, \(r_A(t)=r_B(t)\)
yes i get the first piece , and for part b you let them equal each other because they collide
and for the third question, you'll have to find the minimum value of rB|A.
yes a and b are closest together when d = -3(t+3)i+(vt+6)j is minimized eg D'=0 ???
thank you very much , you were great help and i understand it so much better !!