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The displacement x of a particle at time t is given by $$x = 5 ext{sin}4t + 12 ext{cos}4t.$$ What is the maximum velocity of the particle? (A) 13 (B) 28 (C) 52 (D) 68 - HSC - SSCE Mathematics Extension 1 - Question 7 - 2016 - Paper 1

Question 7

The-displacement-x-of-a-particle-at-time-t-is-given-by--$$x-=-5-ext{sin}4t-+-12-ext{cos}4t.$$---What-is-the-maximum-velocity-of-the-particle?--(A)-13--(B)-28--(C)-52--(D)-68-HSC-SSCE Mathematics Extension 1-Question 7-2016-Paper 1.png

The displacement x of a particle at time t is given by $$x = 5 ext{sin}4t + 12 ext{cos}4t.$$ What is the maximum velocity of the particle? (A) 13 (B) 28 (C) 52... show full transcript

Worked Solution & Example Answer:The displacement x of a particle at time t is given by $$x = 5 ext{sin}4t + 12 ext{cos}4t.$$ What is the maximum velocity of the particle? (A) 13 (B) 28 (C) 52 (D) 68 - HSC - SSCE Mathematics Extension 1 - Question 7 - 2016 - Paper 1

Step 1

Finding the Velocity

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Answer

The velocity of the particle is the derivative of the displacement with respect to time. Therefore,

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(5\text{sin}4t + 12\text{cos}4t).$

Using the chain rule:

$v(t) = 5 \cdot 4\text{cos}4t - 12 \cdot 4\text{sin}4t = 20\text{cos}4t - 48\text{sin}4t.$

Step 2

Finding the Maximum Velocity

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Answer

To find the maximum velocity, we can express the velocity in the form of a resultant amplitude:

$v(t) = R\text{sin}(4t + \phi),$

where

$R = \sqrt{(20^2 + (-48)^2)} = \sqrt{400 + 2304} = \sqrt{2704} = 52.$

Thus, the maximum velocity of the particle is 52.

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