v → = ( 6 t − 4 t 2 ) i ^ − 8 j ^. Here v → is in meters per second and t in seconds.(a) What are the position, speed, and acceleration of the particle when t = 4.3 s?(b) When (if ever) is the acceleration zero? Prove your answer.(c) When (if ever) is the velocity zero? Prove your answer. **Problem Statement:**The velocity of a particle of mass 50 g moving in the xy plane is given by\[\vec{v} = (6t - 4t^2) \hat{i} - 8 \hat{j}\]Here $\vec{v}$ is in meters per second and $t$ in seconds.**Questions:**(a) What are the position, speed, and acceleration of the particle when $t = 4.3$ s?(b) When (if ever) is the acceleration zero? Prove your answer.(c) When (if ever) is the velocity zero? Prove your answer.

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Description: v → = ( 6 t − 4 t 2 ) i ^ − 8 j ^. Here v → is in meters per second and t in seconds.(a) What are the position, speed, and acceleration of the particle when t = 4.3 s?(b) When (if ever) is the acceleration zero? Prove your answer.(c) When (if ever)

Answered: v → = ( 6 t − 4 t 2 ) i ^ − 8 j ^.…

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

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v → = ( 6 t − 4 t 2 ) i ^ − 8 j ^. Here v → is in meters per second and t in seconds.

(a) What are the position, speed, and acceleration of the particle when t = 4.3 s?

(b) When (if ever) is the acceleration zero? Prove your answer.

**Problem Statement:**

The velocity of a particle of mass 50 g moving in the xy plane is given by

\[
\vec{v} = (6t - 4t^2) \hat{i} - 8 \hat{j}
\]

Here $\vec{v}$ is in meters per second and $t$ in seconds.

**Questions:**

(a) What are the position, speed, and acceleration of the particle when $t = 4.3$ s?

(b) When (if ever) is the acceleration zero? Prove your answer.

(c) When (if ever) is the velocity zero? Prove your answer.

Expert Solution

Step 1

Acceleration is defined as the rate of change of velocity with time. So simply by differentiating the expression fir velocity with respect to time we can find out the acceleration. Similarly by integrating the velocity we calculate the position.

a) $\int_{0}^{x} d x = \int_{0}^{t} (6 t - 4 t^{2}) \overset{⏜}{i} d t - \int_{0}^{t} 8 \overset{⏞}{j} d t$

$\overset{⇀}{x} = [3 t^{2} - \frac{4}{3} t^{3}] \overset{⏞}{i} - 8 t \overset{⏞}{j}$

putting t=4.3 s

$\overset{⇀}{x} = - [(50.27) \hat{i} + 34.4 \hat{j]}$

Similarly acceleration ,

$\overset{⇀}{a} = \frac{d (6 t - 4 t^{2}) \hat{i}}{d t} - \frac{d (8 \hat{j})}{d t}$

$\overset{⇀}{a} = (6 - 8 t) \hat{i}$

putting t=4.3 s

$\overset{⇀}{a} = - 28.4 \hat{i}$

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