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1 month ago

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An object has a position given by r⃗ = [2.0 m + (5.00 m/s)t] i^ + [3.0 m − (3.00 m/s2)t2] j^ , where quantities are in SI units.

What is the speed of the object at time t = 2.00 s?

1 answer:

Zina [12.3K]1 month ago

8 0

Answer:Speed = 13.0 m/s

Step-by-step explanation:

Considering velocity, v = (ai ± bj)m/s.

Since speed is a scalar and has only magnitude, we can derive it from the velocity by calculating just the magnitude represented as:

...Equation 1

$Speed = (\sqrt{a^{2} + b^{2} } )m/s$ From the context, the position vector is denoted as

r

To determine the velocity, we differentiate r with respect to

time, t

We know that differentiating a constant results in 0.

$v = \frac{dr}{dt} = \frac{d}{dt}[(2.0+5.00t)i + (3.0 - 3.00t^{2})j] \\\\v = 0 + 5.00i + 0 - (6.00t)j$

At time t = 2.00s,

$v = 5.00i - (6.00*2.00)j = 5.00i - 12.00j$

Using equation 1, we find the speed at this moment to be 13.00m/s

$Speed = \sqrt{(5.00)^{2} + (12.00)^{2} } = \sqrt{25 + 144} \\ \\Speed = \sqrt{169} = 13.00m/s$

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Keep in mind that

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we will now validate each point

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$0=0$ -------> is valid

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the point $(0,0)$ lies on the direct variation line

case B) point $(2.5,6)$

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Insert the values of x and y into the direct variation equation

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case C) point $(3,10)$

$x=3\ y=10$

Insert the values of x and y into the direct variation equation

$10=\frac{12}{5}*3$

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case D) point $(7.5,18)$

$x=7.5\ y=18$

Insert the values of x and y into the direct variation equation

$18=\frac{12}{5}*7.5$

$18=18$ -------> is valid

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the point $(7.5,18)$ lies on the direct variation line

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Insert the values of x and y into the direct variation equation

$24=\frac{12}{5}*12.5$

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the point $(12.5,24)$ does not lie on the direct variation line

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Insert the values of x and y into the direct variation equation

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the point $(15,36)$ lies on the direct variation line

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the solution is

$(0,0)$

$(2.5,6)$

$(7.5,18)$

$(15,36)$

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