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Object A has a position as a function of time given by A(t) = (3.00 m/s)tî + (1.00 m/s2)t2ĵ. Object B has a position as a function of time given by B(t) = (4.00 m/s)tî + (-1.00 m/s2)t2ĵ. All quantities are SI units. What is the distance between object A and object B at time t = 3.00 s?

Respuesta :

Answer:

18.25m

Step-by-step explanation:

Subtracting the two Vectors we have

= 3ti - 4ti + t^2j - - t^2j

=- 1ti + 2 t^2j

= The distance between the lines is given by

= Squaroot of ( ( -3)^2 +( 2(3)^2 )^2

= Squaroot of 9 +324

= Squaroot of 333

= 18.25m

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