Please explain and work out part b with plenty of explanations on the integration parts A sin(wt), where1. An object moves along the x-axis with a velocity given by vx(t)3s¹ and A 1 m/s. (For an example of a physical scenario in which this type ofbehavior might occur, see question 4.)W===(a) What is the displacement Ax of the object between t = = 0 and t=Solve this graphically: Draw a graph of v₂ vs. t for the indicated time range,and determine your answer entirely by examining this graph.2 s?(b) What is the position x(t) of the object as a function of time if x(0) = 0 m?Solve this by direct integration: In other words, solve via x(t) − x(0)So vz(t)dt'.

Answered: What is the position r(t) of the object…

What is the position r(t) of the object as a function of time if x(0) = 0 m? Solve this by direct integration: In other words, solve via x(t) − x( S v₂ (t')dt'.

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Question

Please explain and work out part b with plenty of explanations on the integration parts

A sin(wt), where
1. An object moves along the x-axis with a velocity given by vx(t)
3s¹ and A 1 m/s. (For an example of a physical scenario in which this type of
behavior might occur, see question 4.)
W=
=
=
(a) What is the displacement Ax of the object between t = = 0 and t
=
Solve this graphically: Draw a graph of v₂ vs. t for the indicated time range,
and determine your answer entirely by examining this graph.
2 s?
(b) What is the position x(t) of the object as a function of time if x(0) = 0 m?
Solve this by direct integration: In other words, solve via x(t) − x(0)
So vz(t)dt'.

Expert Solution

Step 1

(b) Given: The velocity of the object is $v_{x} (t) = A \sin (ωt)$ .

The value of $ω$ is $3 s^{- 1}$ .

The value of A is $1 m / s$ .

To determine: The position of the object as a function of time.

The velocity of the object is given as

$v = \frac{dx}{dt}$

where x is the position of the particle.

Substitute the value of v

$\frac{dx}{dt} = A \sin (ωt) dx = A \sin (ωt) . dt$

Integrate this equation on both sides of the equation

$\int_{0}^{x} dx = \int_{0}^{t} A \sin (ωt) . dt x = A \int_{0}^{t} \sin (ωt) . dt x = A \times {[- \frac{\cos (ωt)}{ω}]}_{0}^{t} + C x = - A (\frac{\cos (ωt)}{ω} - \frac{\cos (ω \times 0)}{ω}) + C$

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