SOLUTION: A rocket of mass m = 500 kg is travelling in a straight line for a short time.
The distance in metres covered by the rocket during this time is described
by the function
s(t) =
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-> SOLUTION: A rocket of mass m = 500 kg is travelling in a straight line for a short time.
The distance in metres covered by the rocket during this time is described
by the function
s(t) =
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Question 973015: A rocket of mass m = 500 kg is travelling in a straight line for a short time.
The distance in metres covered by the rocket during this time is described
by the function
s(t) = 220t-15t^2-60ln(t + 1)
where t > 0 is the time in seconds.
(a) Find a function that describes the speed of the rocket.
(b) What is the distance covered by the rocket by time t = 6 seconds?
(c) Find the value of time t when the speed of the rocket is 100 ms^-1
(d) Find a function that describes the acceleration of the rocket
(e) Find the acceleration of the rocket at t = 3 seconds.
(f) Find the time when the rocket's acceleration is -15ms^-2. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A rocket of mass m = 500 kg is travelling in a straight line for a short time.
The distance in metres covered by the rocket during this time is described
by the function
s(t) = 220t-15t^2-60ln(t + 1)
where t > 0 is the time in seconds.
(a) Find a function that describes the speed of the rocket.
Speed is the 1st derivative.
s'(t) = 220 - 30t - 60/(t+1)
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(b) What is the distance covered by the rocket by time t = 6 seconds?
s(t) = 220t-15t^2-60ln(t + 1)
s(6) = 220*6 - 15*36 - 60/7 = 771.43 meters
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(c) Find the value of time t when the speed of the rocket is 100 m/sec
s'(t) = 220 - 30t - 60/(t+1) = 100
Solve for t, ignore the negative value.
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(d) Find a function that describes the acceleration of the rocket
a = 2nd derivative
s"(t) = -30 + 60/(t+1)^2
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(e) Find the acceleration of the rocket at t = 3 seconds.
s"(t) = -30 + 60/16 = -26.25 m/sec/sec
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(f) Find the time when the rocket's acceleration is -15ms^-2.
Not clear.
