The position at time t, in seconds of an object moving along a like is s(t)= 3r^3 - 40.5t^2 + 162t fot 0<= t <=8
When is the obkjection advancing anf whne is it retreating
To determine when the object is advancing or retreating, we need to analyze the velocity of the object. The velocity can be found by taking the derivative of the position equation with respect to time. So, let's find the derivative of the position equation.
s(t) = 3r^3 - 40.5t^2 + 162t
To find the velocity function, we differentiate each term of the position equation with respect to time:
s'(t) = d/dt(3r^3) - d/dt(40.5t^2) + d/dt(162t)
The derivative of a constant term (like 162t) is just the constant itself (in this case, 162).
Differentiating the other terms using the power rule:
s'(t) = 0 - (2 * 40.5t) + 162
Simplifying:
s'(t) = -81t + 162
Now, let's determine when the object is advancing and when it is retreating.
The object is advancing when its velocity (s'(t)) is positive, indicating it is moving in the positive direction.
The object is retreating when its velocity (s'(t)) is negative, indicating it is moving in the negative direction.
So, to find the times when the object is advancing and retreating, we need to find when s'(t) > 0 and when s'(t) < 0.
Setting s'(t) > 0:
-81t + 162 > 0
Adding 81t to both sides:
162 > 81t
Dividing by 81:
2 > t
Therefore, the object is advancing for t < 2.
Setting s'(t) < 0:
-81t + 162 < 0
Adding 81t to both sides:
162 < 81t
Dividing by 81:
2 < t
Therefore, the object is retreating for t > 2.
In summary, the object is advancing for t < 2 seconds and retreating for t > 2 seconds.