Help on a 2016 past paper question (1 Viewer)

narges-ts

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I've been stuck on this for hours:

The displacement x of a particle at time t is given by
x = 5sin4t + 12cos4t .
What is the maximum velocity of the particle?

The answer is 52 but I have no idea how they got it please help!!!
 

jazz519

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Differentiate the equation twice, let that equal 0 and sub the value for t you find that gives a maximum into the first derivative
 

pikachu975

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I've been stuck on this for hours:

The displacement x of a particle at time t is given by
x = 5sin4t + 12cos4t .
What is the maximum velocity of the particle?

The answer is 52 but I have no idea how they got it please help!!!
x = 13sin(4t+tan^-1 (12/5)) using auxiliary angle
x = 0 for max speed
4t+tan^-1 (12/5) = 0
t = -tan^-1 (12/5) / 4

v = 20cos4t - 48sin4t
sub t in
v = 52

Edit: Integrand's method is faster
 
Last edited:

InteGrand

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I've been stuck on this for hours:

The displacement x of a particle at time t is given by
x = 5sin4t + 12cos4t .
What is the maximum velocity of the particle?

The answer is 52 but I have no idea how they got it please help!!!
 
Last edited:

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