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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? - HSC - SSCE Mathematics Extension 1 - Question 7 - 2016 - Paper 1

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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?-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?

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? - HSC - SSCE Mathematics Extension 1 - Question 7 - 2016 - Paper 1

Step 1

Find the velocity function

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Answer

The velocity of the particle is the derivative of the displacement function with respect to time. Therefore, we compute the derivative:

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

Using the chain rule, this becomes:

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

Step 2

Find the maximum velocity

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Answer

To find the maximum velocity, we need to determine the maximum value of the velocity function:

v(t)=20cos(4t)−48sin(4t).v(t) = 20\text{cos}(4t) - 48\text{sin}(4t).

Using the amplitude formula for a function of the form Acos+BsinA\text{cos} + B\text{sin}, we can find the maximum velocity as:

Vmax=(20)2+(−48)2=400+2304=2704=52.V_{max} = \sqrt{(20)^2 + (-48)^2} = \sqrt{400 + 2304} = \sqrt{2704} = 52.

Thus, the maximum velocity of the particle is 52.

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