how do i do this word problem?

Question

The function d(t)=8t^2 +16, models the distance in km that a motorcycle travels in t hours.

Determine the motorcycle’s average speed from a time of 2 hours to a time of 4 hours and

compare that to its instantaneous speed at 3 hours.

Answer 1

To find average speed over a finite time interval using the distance (position) function, we can simply calculate the distance traveled during that time and divide by the length of the time interval. The calculation looks like this:

d(4) = 144 km , d(2) = 48 km , avg speed = (144 - 48) km / (4 - 2) hrs = 48 km/hr

If this calculation reminds you of calculating slope, (y₂ - y₁) / (x₂ - x₁) , that's good -- it should. The average rate of change of any function is simply the slope of the secant line connecting two points on the graph of the function. In this case, the 2 points are (144 , 4) and (48 , 2). And speed is the average rate of change of position.

Instantaneous speed at t = 3 is found by taking the first derivative and plugging in t = 3 as follows:

d^'(t) = 16t

d^'(3) = 48 km/hr

Answer 2

This problem asks you to do two computations:

1. average speed of the motorcycle from time t=2 to time t=4
2. instantaneous speed at the middle of this interval, at time t=3

For the first of these two, we're looking at concepts from prior to calculus class: we just need the relationship that distance = speed * time. Since in this case we want to solve for speed, we use algebra to isolate that piece:

speed = distance/time

We know the start and end times - 2 hours and 4 hours - so the total time travelled is just the difference of these, 4-2

We can likewise find the start and end distances without too much trouble, by plugging into the given function at each of the desired times:

d(2) = 8*2^2 +16 = 48

d(4) = 8*4^2 +16 = 144

for a total distance travelled of 144-48 miles.

Our final speed, then is this change in distance over change in time: (144-48)/(4-2) = 48 miles/hour.

I have been leaving things in terms of the changes instead of fully simplifying each step, because this should look somewhat suggestive: this average speed in miles per hour is exactly the change in our y-value of miles over the change in our x-value of hours... it's the slope from point (2,d(2)) to point (4,d(4)) on the graph!

b) The above leads nicely into why we'd want to use a derivative to compute the instantaneous rate of change; we want the slope no longer of a secant line between two points of the graph, but rather of one tangent at exactly one point... namely, (3,d(3)).

I'm not sure where you are on your calculus journey yet; perhaps you know the power rule and are ready to jump strait from

d(t) = 8t^2 +16

to

d'(t) = 8*2t + 0 = 16t

and from there know that d'(3) = 16*3 = 48 miles/hour

BUT IF NOT! it's totally fine, we can set up a very slope-formula-like approach to this and take a limit to get the same answer. Take the point (3,d(3)) and an additional point some small distance h away, (3+h,d(3+h)). We're going to set up the same slope formula of (change in y)/(change in x) and then see what happens as h gets super small. First, let's actually compute those y-values, then we can plug them into the slope fomula:

d(3) = 8*3^2 +16 = 88 miles

d(3+h) = 8*(3+h)^2 +16 = 8*(3^2+6h+h^2)+16 miles, which I'm not even going to bother to simplify because actually things are gonna work out nicer algebraically if I leave it as-is. Calc pro tip: don't simplify midway through unless it's toward a specific, known goal, otherwise you're just spending extra time practicing your algebra, potentially getting things farther away from their most-useful representation for what's coming next, and leaving room for introducing a mistake =P

Cool, I've got points (3,8*3^2 +16) and (3+h,8*(9+6h+h^2)+16). The slope between them is found by taking the change in the y-values and dividing by the change in x-values:

8*(3^2+6h+h^2)+16 - (8*3^2 +16)

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

3+h - 3

Ok, now I am gonna simplify, because ideally I'd like to know what this becomes as h gets small. I notice immediately that there's a +16 and a -16 in the numerator, and a +3 and -3 in the denominator; cancel these inverses to get

8*(3^2+6h+h^2) - 8*3^2

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

h

Our end goal is to let h get small, but right now with that denominator that means we're gonna be dividing by smaller and smaller values, which can get weird... the limit of that process is division by 0, which is a big no-no for mathematicians! The good news is, we're gonna be able to use algebra to get ourselves out of the problem, here and in most of these set-ups.

I see an 8*3^2 and a -8*3^2 - let's go ahead and distribute that 8* over the parentheses, in order to let those additive inverses cancel out:

8*6h+8*h^2

----------

h

ok, awesome, we now are in a place where everything in the numerator has an h in it! This means we can get rid of the h in the denominator by cancelling it with a copy of h in each piece of the numerator:

8*6 + 8*h is our fully-simplified formula for the slope between (3,d(3)) and the very-nearby point (3+h,d(3+h)).

As h gets closer and closer to 0, 8*6+8h gets closer and closer 8*6+8*0 = 8*6 = 48 miles/hour! This is our instantaneous rate of change at t=3 hours.

Coincidentally, this is also the average rate of change between t=2 and t=4 hours! This will not always be the case, but it happens to work out nicely for evenly-spaced intervals with this function. There's an algebraic approach for seeing why that's the case, too, but that's probably worth a question of its own :)

I hope this was helpful! Please feel free to leave a comment with a follow-up question if needed.

Answer 3

Using the definition of average value from calculus, we can write

Average speed = ( ∫ modulus of dx from t₁ to t₂ ) / t₂ - t₁ = total distance / total time Δt.

In our case

v _avr = (d(t₂) - d(t₁) ) / t₂ - t₁ = ( (8·(4)² + 16) - (8·(2)² + 16)) / (4 - 2) = 48 km/h.

Instantaneous speed v we will find, if take the derivative of distance with respect to time. So, we have

v(t) = d(8t² + 16)/dt = 16t

Substituting 3 for time t, we finally have

v(3) = 16·3 = 48 km/h

In both cases we got the same answer 48 km/h

This fact is easy to explain from physics point of view. In our case we have uniformly accelerated motion with acceleration a = 16 m/s². Only for uniformly accelerated motion average speed is the same as average arithmetic (mean). If the time interval between 2 and 4, then average speed for this time interval is the same as instantaneous speed at 3 units of time.