8. A particle is moving in a straight line such that the relation below gives its position. Calculate (a) its2.00 s to t₂ = 5.00 s, and (b) its instantaneous=average acceleration during the time interval from t₁acceleration as a function of time.x=(7.2 m/s²)² + (0.3 m/s)+(2.1m)

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Description: 8. A particle is moving in a straight line such that the relation below gives its position. Calculate (a) its2.00 s to t₂ = 5.00 s, and (b) its instantaneous=average acceleration during the time interval from t₁acceleration as a function of time.x=(

position as a function of time is given as x(t)=7.2 m/s2×t2 + 0.3 m/s×t+2.1 mvelocity is defined as…

Answered: 8. A particle is moving in a straight…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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8. A particle is moving in a straight line such that the relation below gives its position. Calculate (a) its
2.00 s to t₂ = 5.00 s, and (b) its instantaneous
=
average acceleration during the time interval from t₁
acceleration as a function of time.
x=(7.2 m/s²)² + (0.3 m/s)+(2.1m)

Expert Solution

Step 1

position as a function of time is given as

$x (t) = (7.2 m / s^{2}) \times t^{2} + (0.3 m / s) \times t + (2.1 m) v e l o c i t y i s d e f i n e d a s t h e r a t e o f c h a n g e o f p o s i t i o n o f t h e p a r t i c l e v (t) = \frac{d x (t)}{d t} \Rightarrow v (t) = (7.2 m / s^{2}) \times (2 \times t) + (0.3 m / s) \times (1) + 0 v (t) = (14.4 m / s^{2}) \times t + (0.3 m / s) a) l e t a_{a v g} d e n o t e s t h e a v e r a g e a c c e l e r a t i o n d u r i n g t h e t i m e i n t e r v a l f r o m t_{1} = 2 s t o t_{2} = 5 s a_{a v g} = \frac{∆ v}{∆ t} w h e r e ∆ v i s t h e c h a n g e i n v e l o c i t y ∆ t i s c h a n g e i n t i m e ∆ t = t_{2} - t_{1} = (5 - 2) s = 3 s$

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