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allochka39001 [22]

3 years ago

13

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 f

unction 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?

1 answer:

Vilka [71]3 years ago

8 0

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

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

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Answer:

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The electric field at x = 3L is 166.67 N/C

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As per the question:

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Electric field, $\vec{E_{2L}} = 500N/C$

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$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

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Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

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Then at x = 3L:

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