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A hiker goes on a three-hour hike and walks in a straight line towards
a lake which is five miles away from her when she starts walking.

suppose that her velocity is given by $$v = v(t) = -3 t^3 + 12t^2 +9t.$$

1. Does the hiker ever reach the lake? (I think no since it is going below the axis after $t=3$.)




2. How close does she get to the lake? (No idea. Please help.)




3. When is she farthest from the lake? How far is she from the lake? (Again, no clue.)




4. When is she going the fastest? just find the max of velocity graph which is t=2.215




5. When did she switch directions? (Just factor it and set it equal to zero. $t=0$, $t=1$, $t=3$.)




6. What was the total distance she hiked? 8+1.25=9.25

Velocity is the derivation of her trajectory (position) by time, so her position in time $t$ is

$$x(t) =int_0^tv(t),dt = -frac34t^4+4t^3+frac92t^2 = t^2left(-frac34t^2+4t+frac92right)$$

From you question I suppose that time is in hours and position in miles.

Now you may use this expression to answer your questions:

1. $x(3) = frac3514 = 87.75 > 5 text(miles)$ - it seems that she drowned in the lake!




2. See 1. - so close, so she drowned in the lake! (Something is wrong with time/distance units or with equation in your question.)
   
   (Need to obtain maximum of $x(t)$ for $t in [0,3]$, and subtract it from her initial 5 miles distance from the lake.)




3. (Need to obtain minimum of $x(t)$ for $t in [0,3]$, and subtract it from her initial 5 miles distance from the lake.)




4. See 3. - in the time in [0,3], when she reach her longest distance from the lake.

The conclusion:

As I wrote in 2., something is wrong with the task in your question, because the results are strange. Please check it.

Assuming the correct expression for the velocity function should be $ v(t) = -3t^3+12t^2-9t $ then:

1. Your answer to this is correct, but given that the hiker starts walking towards the lake ($ v^prime(0) < 0 $, so $ v(t) < 0 $ for $ t $ just after zero), you need to check that the hiker doesn't reach the lake before she turns around and starts walking away from it, which she does when the velocity function changes sign at $ t = 1 $. Substituting this value of $ t $ into the position function, will tell you how far she is from the lake at that time (which I find to be a positive distance). It would appear from your answer to item 6 that you have actually done this.


2. See answer to part 1. Given that the hiker continues to walk away from the lake until $ t=3 $, the closest she gets is when she turns around at $ t = 1 $.


3. Since the hiker continues to walk away from the lake until $ t=3 $, the farthest she gets from the lake is either the $5$ miles she is at the start, or the distance she is at $ t=3 $, whichever is the larger. So you have to substitute $ t=3 $ into the distance function to find that out.


4. I get the same answer for this as you did ($ t=frac8+sqrt286$ $approx2.215 $).


5. I agree that the hiker switches directions at $ t=1 $, but I wouldn't say she does so at $ t=0 $ when she merely starts walking, or at $ t=3 $ when she stops.


6. I obtained the same answer for this as you did.