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Object A has a position as a function of time given by A(t) = (3.00 m/s)t ˆı + (1.00 m/s2 )t2 ˆ. Object B has a position as a function of time given by B(t) = (4.00 m/s)t ˆı + (-1.00 m/s2 )t2 ˆ. All quantities are SI units. What is the distance between object A and object B at time t = 3.00 s?

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lublana

Answer:

18.25 units

Explanation:

We are given that

The position of object A=[tex]A(t)=(3.00m/s)ti+(1.00m/s^2)t^2j[/tex]

The position of object B=[tex]B(t)=(4.00m/s)ti+(-1.00m/s^2)t^2j[/tex]

We have to find the distance between object A and object at time t=3.00 s

Substitute t=3 s

[tex]A(3)=9i+9j[/tex]

[tex]B(3)=12i-9j[/tex]

The distance between A and B

[tex]A(3)-B(3)=9i+9j-12i+9j=-3i+18j[/tex]

The magnitude of distance between A and B

[tex]\mid r\mid=\sqrt{(-3)^2+(18)^2}=18.25 units[/tex]

Using the formula magnitude of position vector

r=xi+yj+zk

[tex]\mid r\mid=\sqrt{x^2+y^2+z^2}[/tex]

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