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Calculus question on Position, Velocity, Acceleration. Easy 10 pts! Please include explanation/steps

Author : ?

Submitted : 2018-06-14 04:53:48    Popularity:     

Tags: Velocity  Acceleration  Position  Calculus  question  

A particle moves along the x-axis so that its position at time t, 0≤t≤5, is given by the function s(t)= t^3-6t^2+9t+6. When is the speed of the particle decreasing?

Answers:

v = s'(t) = 3t^2-12t+9
a = s"(t) = 6t-12 < 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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