x(t)=3t^3-40.5t^2+162t find where object is speeding up or slowing down
x ' (t) = 9t^2 - 81t + 162
so objects are "speeding up" when x ' (t) > 0
and slowing down when x ' (t) < 0
Consider the zeros of the expression
9t^2 - 81t + 162 = 0
t^2 - 9t + 18 = 0
(t-3)(t-6) = 0
t = 3 or t = 6
so the t of the vertex would be half-way between those values or t = 4.5
Using the properties of the quadratic and knowing that x ' (t) is a parabola opening upwards
the object is speeding up for t > 4.5 seconds
and is slowing down for t < 4.5 seconds
Here is a graph of the velocity
http://www.wolframalpha.com/input/?i=plot+y+%3D+9t%5E2+-+81t+%2B+162
To determine where the object is speeding up or slowing down, we need to find the critical points of the velocity function.
The velocity function is the derivative of the position function, so we need to differentiate x(t) to find the velocity function:
v(t) = d/dt (x(t)) = d/dt (3t^3 - 40.5t^2 + 162t)
Taking the derivative:
v(t) = 9t^2 - 81t + 162
To find the critical points, we need to solve for when the velocity is equal to zero:
9t^2 - 81t + 162 = 0
Dividing the equation by 9:
t^2 - 9t + 18 = 0
Now we can factorize the equation:
(t - 3)(t - 6) = 0
Setting each factor equal to zero:
t - 3 = 0 => t = 3
t - 6 = 0 => t = 6
So the critical points are t = 3 and t = 6.
To determine whether the object is speeding up or slowing down at these points, we need to analyze the sign of the velocity function before and after the critical points.
For t < 3, substitute a value less than 3 into the velocity function:
v(t) = 9t^2 - 81t + 162
v(2) = 9(2)^2 - 81(2) + 162 = 18 - 162 + 162 = 18
Since v(2) > 0, the object is speeding up before t = 3.
For 3 < t < 6, substitute a value between 3 and 6 into the velocity function:
v(t) = 9t^2 - 81t + 162
v(4) = 9(4)^2 - 81(4) + 162 = 144 - 324 + 162 = -18
Since v(4) < 0, the object is slowing down between t = 3 and t = 6.
For t > 6, substitute a value greater than 6 into the velocity function:
v(t) = 9t^2 - 81t + 162
v(7) = 9(7)^2 - 81(7) + 162 = 441 - 567 + 162 = 36
Since v(7) > 0, the object is speeding up after t = 6.
In conclusion, the object is speeding up before t = 3 and after t = 6, and it is slowing down between t = 3 and t = 6.
To determine where the object is speeding up or slowing down, we need to find the values of t where the acceleration is positive (speeding up) or negative (slowing down).
The acceleration is the second derivative of the position function, x(t). To find it, we need to take the derivative of x(t) twice.
First, let's find the first derivative of x(t):
x'(t) = d/dt (3t^3 - 40.5t^2 + 162t)
= 9t^2 - 81t + 162
Now, let's find the second derivative of x(t):
x''(t) = d/dt (9t^2 - 81t + 162)
= 18t - 81
To find where the object is speeding up or slowing down, we need to find the values of t where x''(t) is positive or negative.
Setting x''(t) greater than 0 to find where the object is speeding up:
18t - 81 > 0
18t > 81
t > 81/18
t > 4.5
Setting x''(t) less than 0 to find where the object is slowing down:
18t - 81 < 0
18t < 81
t < 81/18
t < 4.5
Therefore, the object is speeding up for t > 4.5 and slowing down for t < 4.5.