Minimum Speed

Question

An everglades can accelerate at 2m/sec². A pilot wnats to do 2 laps of a 600 meter course with an average speed of 20 m/sec. If he does the 1^st lap at a constant 15m/sec, what is the minimum speed he should accelerate to (and hold) to finish the course?

Answer 1

To achieve an overall average speed of 20m/s, the pilot realizes that it needs to take 30 seconds ( delta t = delta x / speed = 600 m / (20m/s)) to finish each 600 meter lap, so two laps should take precisely one minute.

 

But if the first lap is completed at 15m/sec, that would have used up 40 seconds (delta t = delta x / speed) , leaving the pilot only 20 seconds to complete the last lap.

 

We know that the pilot must accelerate or he'll end up using another 40 seconds and complete the race 20 seconds late.  To accelerate the least, he needs to start the acceleration immediately, hold to the maximum acceleration allowed, and maintain the final speed for the rest of the way.

 

Thinking about the velocity versus time graph, what would the pilot's intended motion look like?  Form an answer before you read on.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

If you said, "a straight line with a positive slope of (2 meters per square second), followed by a flat line representing the final (and maximum) speed", you were correct!

 

So here's another question: How does the distance travelled relate to a velocity versus time graph?  Think of an answer before you read on.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I bet you said "The displacement in any time period is equal to the area under the velocity versus time graph".

 

Ok, what is the area under the shape you mentioned above?  We need a formula for it and we need the area to be equal to the remaining 600 m while the width of the shape can only be 20 seconds.  We know the initial velocity for this lap is 15m/s, so we could think of the lower part of the shape as a rectangle that is 15 m/s high and 20 seconds wide.  Area of a rectangular shape equals length times width.  The "length" of this rectangle on our graph is its height along our velocity axis (15m/s).  The width is the change in time (20 seconds for the second lap).  So the area is 15 m/s times 20 s = 300 m.

 

We also have to add to that the upper portion of the area of the graph. This could be thought of as a trapezoid. The area of a trapezoid is the average length of the two parallel sides, times the perpendicular dimension width.  In this case, we know the long length t_long is (20 seconds) and we know what the area needs to be (600 m). We don't know the short side L_short.

A_trapezoid [m] = ((t_long + t_short)[s])*((v_final - 15) m/s) / 2

 

We know that ( v_final - 15 m/s ) represent the rise and the slope is 2 (m/s)/s. Since slope = rise/run, run = rise/slope.

Therefore the run is ( ((v_final - 15) m/s)/ ((2 m/s)/s) ) =  ( 0.5 v_final - 7.5 ) s which is the difference in lengths.

Therefore the short side length t_short = t_long - run = 20 - (0.5 v_final - 7.5) seconds

 

Now we can substitute this expression for L_short into the area of a trapezoid and solve for v_final.

 

A_trapezoid = (20 + (20 - (0.5 v_final - 7.5)))[s]* (v_final - 15)[m/s] = 600 [m]]

 

(The units do work out so we can cancel and stop regarding them, noting that v_final will have units of [m/s]).

 

Simplifying: 

(47.5 - 0.5 v_final)* (v_final - 15) = 600

This is going to be a quadratic, so we are going to have to set everything equal to zero.

 

Foiling out the first two factors using a factoring box first (v now represents v_final)

               v                  - 15

         ===================

47.5   |   95/2 v        |   -15(95)/2  |

         ----------------------------------

-0.5v  |  - (v^2) / 2  |  15 v / 2      |

        ===================

 

-(v^2) + 55 v - (15)*(95) - 600 = 0

 ------                --------

   2                       2  

 

Multiply everything by -2 to normalize the first term:

 

v^2 - 110 v + 2625 = 0

 

v = - b ±√(b²-4ac)

       ----------------

            2a

 

   =  -(-110) ±√( (-110)² - 4*(1)*(2625) )

         ----------------------------

                2*(1)

 

(Note: we can see the square root term is going to be bigger than 110, so we'll get a positive answer and a negative answer, but we can be sure that we should reject the negative answer in this problem)

 

=    110 ± √( 1600  )
    ----------------------------
                2*(1)

 

=  55 ± 20

 

Check bigger answer:

55+20 m/s = 75 m/s

BUT Check: How long would it take to accelerate from 15 m/s to 75 m/s at (2 m/s)/s?

60 m/s / ((2 m/s)/s) = 30 seconds.

That's not going to work! The pilot doesn't have 30 seconds left in this race!

 

And check smaller answer:

55 - 20 = 35 m/s

 

That's 20m/s faster then when he started the second lap, and it would take ten seconds (delta t = delta v / acceleration) to do which is half the time remaining at the end of the first lap, so it works!

 

Final answer:  He needs to accelerate to and hold a final speed of 35 m/s.

 

 

Answer 2

Hi Tiffany!

 

I will assume we are talking about a swamp boat, and that is must accelerate at 2 m/s².  You can actually show a larger acceleration associates with the minimum speed asked for in the second part, so that the largest acceleration possible is related to the minimum speed, but that is more involved and uses calculus (at least the way I demonstrated it).

 

So, with those assumptions, we can set up the problem as follows:

 

The desired average speed is 20 m/s.  Average speed is defined as total distance covered over total time.  The total distance covered in two laps around this course is 1200 m (600 m x 2).  Therefore, for an average speed of 20 m/s, he needs:

 

v = 20 m/s = 1200 m/t_tot

 

This means t_tot = 60 s for him to finish the two laps.

 

We are also told he runs the first lap at a constant 15 m/s.  We can calculate the time it takes to run the first lap:

 

600 m = (15 m/s)t₁ -->  t₁ = 600/15 = 40 s

 

So the first lap takes 40 s, and he needs to take a total of 60 s for both laps (to get the desired average speed).  We thus have two main restrictions on lap 2:  that the distance covered is 600 m, and the time is 20 s.

 

In general, then, we can consider a possibility (as the problem suggests) where the boat accelerates (at 2 m/s²) for some time in lap 2, then reaches and maintains a constant speed the rest of the time.  We can thus break lap 2 into two parts:

 

Part a:

 

time = t_2a

acceleration = a = 2 m/s²

initial velocity = v_oa = 15 m/s (from the end of lap 1)

 

With these values, we can determine expressions for the final velocity (v_a) and displacement (x-x_o)_a of this segment:

 

v_a = v_oa + at_2a = 15 m/s + (2 m/s²)t_2a = 15 + t_2a

(x-x_o)_a = v_oat_2a+(1/2)at_2a² = 15t_2a+(1/2)(2)t_2a² = 15t_2a+t_2a²

 

Then, we can look at Part b of lap 2, where the boat would maintain a constant speed:

 

time = t_2b

velocity (speed) = (15 + 2t_2a) (it maintains the final speed of Part a)

(x-x_o)_b = velocity * time (for constant speed) = (15+2t_2a)t_2b

 

Now, these equations end up having two unknowns: t_2a and t_2b.  And, for those two unknowns, we can make two relationships, based on the circumstances:

 

1.  (x-x_o)_a+(x-x_o)_b = 600 m  (the total displacement over a and b must be the 600-m distance of lap 2)
2.  t_2a+t_2b = 20 s  (to get the desired average speed

 

We can put these together into one equation.  I choose to solve for t_2a, so I rearrange equation 2 first:

 

t_2b = 20-t_2a

 

(x-x_o)_a + (x-x_o)_b = 600

 

[using our expressions for the displacements in Part a and Part b in terms of times from above]

 

15t_2a + t_2a² + (15 + 2t_2a)t_2b = 600

15t_2a + t_2a² + (15+2t_2a)(20 - t_2a) = 600

 

I have this simplifying to the following quadratic equation in t_2a:

 

t_2a2-40t_2a+300 = 0

 

(which can actually be factored)

 

This gives two solutions for t_2a, only one of which meets the criterion that it must be less than 20 s.  Once you solve for this value of t_2a, you can use it to solve for the velocity the boat attains once it is done accelerating in Part a of lap 2.  I ended up with a speed of 35 m/s.

 

I hope this helps, but there is a lot here, so if you have any questions, just let me know.  I will also check my work again later and see if there are any mistakes, and let you know if I find one.