# Thread: Help with derivative application and application in relation to time.

1. ## Help with derivative application and application in relation to time.

I have a Calculus test tomorrow, and I'm honestly lost here. This is from a sample test.

5. The function s(t)= -2t^3 + 21t^2 - 60t + 52 gives the distance (inches) of an object from a certain fixed point as a function of time (seconds). Assume the object is moving in one dimension (which means a straight line path…i.e. this is rectilinear motion).

a) Does the object ever reach the “certain fixed point” (if so tell when)? Answer yes or no and describe how you got your answer. (6 points)

I have no idea how to figure this part out.

b) Let’s say the object is advancing when it’s moving towards the fixed point and retreating when it’s moving away from the fixed point. Using these definitions, determine time intervals when the object is advancing and retreating. (8 points)

Derivative is used here, if I'm not mistaken.
s(t)= -6t^2 + 42t - 60

No idea where to go from there.

c) Let’s say acceleration happens when speed increases and deceleration happens when speed decreases. Using these definitions, determine time intervals when the object is accelerating and decelerating.
(8 points)

Second derivative here, I'm pretty sure. No idea what to do with it.
s(t)= -12t + 42

I need some serious help here.

2. Part a.
The equation tells you the distance to the "certian fixed point". Therefore they are asking you if there is a solution to the equation:

$\displaystyle s(t)=0= -2t^3 + 21t^2 - 60t + 52$

Now if this can be solved then there will be values of t for which s(t) is negative. I suggest finding the values of t for which s(t) is maximum or minimum (the turning points). That is the point at which the derivative is equal to zero ($\displaystyle s(t)=0= -6t^2 + 42t - 60$ or $\displaystyle 0= t^2 + -7t + 10$)

Then when you know the value of t at the turning points substitute them into the equation for s(t) to find the extreme values of s(t). If they are all positive (or all negative) then the answer to their question is "no".

Part b.
The values that you found in part a define the boundaries between where s(t) is advancing and when it is retreating. You can tell if it is advancing or retreating from the value of s'(t). Unless I am mistaken a positive value means retreating.

Part c.
s(t)= -12t + 42
When t > 42/12 s''(t) is negative (accellerating towards the certian fixed point).
When t< 42/12 s''(t) is positive (decellerating towards the fixed point.

I could have my signs back to frot in here so if you think I am wrong, then you may be right!

### advancing and retreating calculus

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