**Question:**A particle moves along a straight line such that its displacement at any time \( t \) is given by \[ x = (t^3 - 6t^2 + 3t + 4) \, \text{m} \]The velocity when the acceleration is zero is:- \( 3 \, \text{ms}^{-1} \)- \(-12 \, \text{ms}^{-1} \)- \( 42 \, \text{ms}^{-1} \ )- \(-9 \, \text{ms}^{-1} \)**Solution Explanation:**To determine the velocity when the acceleration is zero, we need to perform the following steps:1. **First Derivative (Velocity):** Find the first derivative of the displacement function \( x(t) \) to get the velocity function \( v(t) \). \[ v(t) = \frac{dx}{dt} = 3t^2 - 12t + 3 \]2. **Second Derivative (Acceleration):** Find the second derivative of the displacement function \( x(t) \) to get the acceleration function \( a(t) \). \[ a(t) = \frac{d^2x}{dt^2} = 6t - 12 \]3. **Set Acceleration to Zero:** Set the acceleration function \( a(t) \) to zero and solve for \( t \). \[ 0 = 6t - 12 \] \[ t = 2 \]4. **Calculate Velocity at \( t = 2 \):** Substitute \( t = 2 \) into the velocity function \( v(t) \). \[ v(2) = 3(2)^2 - 12(2) + 3 \] \[ v(2) = 3(4) - 24 + 3 \] \[ v(2) = 12 - 24 + 3 \] \[ v(2) = -9 \, \text{ms}^{-1} \]Therefore, the velocity when the acceleration is zero is \(-9 \, \text{ms}^{-1} \).

Given, Position as a function of time = x(t) = t3 - 6t2 + 3t + 4 From the position as a function…

A particle moves along a straight line such that its displacement at any time t is given by x = (t³ - 6t² + 3t + 4) m The velocity when the acceleration is zero is O 3 ms-1 O-12 ms-1 42 ms-1 O-9 ms-1

A particle moves along a straight line such that its displacement at any time t is given by x = (t³ - 6t² + 3t + 4) m The velocity when the acceleration is zero is O 3 ms-1 O-12 ms-1 42 ms-1 O-9 ms-1

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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