Calculus question on Position, Velocity, Acceleration. Easy 10 pts! Please include explanation/steps
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Submitted : 20180614 04:53:48 Popularity:
Tags: Velocity Acceleration Position Calculus question
A particle moves along the xaxis so that its position at time t, 0≤t≤5, is given by the function s(t)= t^36t^2+9t+6. When is the speed of the particle decreasing?
v = s'(t) = 3t^212t+9
a = s"(t) = 6t12 < 0
Answer t < 2
.
Speed, v(t), is the first derivative of the given function.
v(t) is decreasing where its derivative v’(t) < 0 . This is deceleration.
s(t) = t³  6t² + 9t + 6
v(t) = 3t²  12t + 9
v’(t) = 6t  12
6t  12 < 0
t < 2
∴ 0 < t < 2
━━━━━━➤ decreasing on the interval 0 < t < 2
Speed of the particle = d(s)/dt = 3 t²  12 t + 9
The speed will start decreasing after it has achieved a velocity equal to zero.
Velocity = 0
ie 3 t²  12 t + 9 = 0
ie 3 t²  9 t  3 t + 9 = 0
=> 3 t ( t  1 )  3 ( t  1 ) = 0
=> ( 3 t  3 ) ( t  1 ) = 0
Hence velocity will be zero when t = 1 s.
Hence after 1 s the velocity will decrease. ......... Answer Answer
The velocity v(t) is the derivative of the position. So, evaluate v(t) = s'(t) = [left for you].
The speed is the magnitude of that (the absolute value of that).
From here, there are many ways to continue. Here is one way: notice that your v(t) is a simple
"parabola", hence you know everything about it (its behavior, zeros etc). Sketch it, and more importantly, its absolute value, since you want  v(t)  . Where (at what times) is this sketch decreasing? Done!
Or, v(t) is decreasing there where v'(t) < 0. But that is often a trap, since v(t) is not the speed; its the velocity.
The speed v(t) is decreasing there where v(t)' < 0. Solve for t. Done again!
Show your steps here if need be and we can take it from there, else we can assume you finished it on your own!
s ' ( t ) < 0 means solving a quadratic...someone using calculus certainly can compute that ...


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