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
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?
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 ( v ) of the particle is the derivative of the displacement with respect to time ( t ):
v=dtdx=dtd(5sin(4t)+12cos(4t))
Using the chain rule, we get:
v=5⋅4cos(4t)−12⋅4sin(4t)=20cos(4t)−48sin(4t).
Step 2
Determine the maximum velocity
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Answer
To find the maximum value of velocity, we can compute the amplitude of the velocity function, which can be expressed in the form:
v=Rsin(θ+ϕ)
Where:
( R = \sqrt{(20)^2 + (-48)^2} )
( \phi = \tan^{-1}\left(\frac{-48}{20}\right) )
Calculating ( R ):
R=400+2304=2704=52.
Thus, the maximum velocity of the particle is ( 52 ).
