Question

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?

Answer 1

Answer:

18.25 units

Explanation:

We are given that

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

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

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

Substitute t=3 s

$A(3)=9i+9j$

$B(3)=12i-9j$

The distance between A and B

$A(3)-B(3)=9i+9j-12i+9j=-3i+18j$

The magnitude of distance between A and B

$\mid r\mid=\sqrt{(-3)^2+(18)^2}=18.25 units$

Using the formula magnitude of position vector

r=xi+yj+zk

$\mid r\mid=\sqrt{x^2+y^2+z^2}$