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2 years ago

8

Me sitting in my 1997 Ford F-150 on August 19th 2026 at 3:37PM with a shotgun in my lap as the skinwalker disguised as my 7 year

old son walks up to the car. (It thinks it can kill me)

Advanced Placement (AP)

2 answers:

geniusboy [140]2 years ago

8 0

Answer:

5

Explanation:

mestny [16]2 years ago

3 0

Answer: don't know sorry

Explanation:

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Diano4ka-milaya [45]

Answer:

no because the forces are acting on different objects

Explanation:

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1 year ago

For time t≥0 , the acceleration of an object moving in a straight line is given by a(t)=sin(t23). what is the net change in velo

Firlakuza [10]

The net change in velocity during the given time interval is equal to -0.665.

<u>Given the following data:</u>

Initial time = 0.75 seconds.

Final time = 2.25 seconds.

Acceleration, $a(t) = sin(\frac{t^2}{3} )$

<h3>How to calculate the net change in velocity.</h3>

Since acceleration is the rate of change of velocity of an object, we would derive an expression for the velocity by differentiating with respect to time:

$V(t) = \frac{d}{dt} a(t)\\\\V(t) = \frac{d}{dt} (sin(\frac{t^2}{3} ))\\\\V(t) = cos( \frac{t^2}{3})(\frac{2t}{3})\\\\V(t) = \frac{2t}{3}cos( \frac{t^2}{3})$

Substituting the time interval, we have:

$V(t) = \frac{2\times 2.25}{3}cos( \frac{2.25^2}{3})- \frac{2\times 0.75}{3}cos( \frac{0.75^2}{3})\\\\V(t) = \frac{4.5}{3}cos( \frac{5.0625}{3})- \frac{1.5}{3}cos( \frac{0.5625}{3})\\\\V(t) =1.5(-0.11643)-0.5(0.98247)$

V(t) = -0.665.

Read more on acceleration here: brainly.com/question/24728358

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