Question

A particle moving along the x-axis has its velocity described by the function vx = 2t^2 m/s, where t is in s. Its initial position is x0 = 2.0 m at t0 = 0 s. A) At 2.7 s, what is the particle's position? B) At 2.7 s, what is the particle's velocity? C) At 2.7 s, what is the particle's acceleration?

          A particle moving along the x-axis has its velocity described by the function vx = 2t^2 m/s, where t is in s. Its initial position is x0 = 2.0 m at t0 = 0 s.

A) At 2.7 s, what is the particle's position?
B) At 2.7 s, what is the particle's velocity?
C) At 2.7 s, what is the particle's acceleration?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A particle moving along the x-axis has its velocity described by the function vx = 2t^2 m/s, where t is in s. Its initial position is x0 = 2.0 m at t0 = 0 s. A) At 2.7 s, what is the particle's position? B) At 2.7 s, what is the particle's velocity? C) At 2.7 s, what is the particle's acceleration?
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Transcript

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00:01 Hi there, here particle is moving along x -axis and its velocity is described by the function to t squared meters per second.
00:16 So at the initial position when time is zero seconds, x0 is 2 meters.
00:33 And here we have to calculate position at 2 .7 seconds.
00:48 Let's do this.
00:53 So here, the position of the particle equals to the integral of its velocity.
01:11 And here that is integral from 0 to time t, which is which equals to integral of 2t squared d t.
01:35 So let's actually here let's calculate the indefinite integral for now for now.
01:45 We'll convert it to the definite integral.
01:48 So the anti -derivative of this function 2d squared is 2 thirds t -ccccc and plus constant.
02:06 So when time equals to 0 x equals to 2, that's why the function is 2 thirds ccube plus 2 .0 that is in meters.
02:25 So therefore when x equals to 2 .7 seconds x equals to 2 thirds times 2 .7 seconds cubed plus 2 .0 and it's all in meters.
02:50 Let's calculate it...
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