Antiderivative word problem --> Acceleration to velocity to position

Question

Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 m/s and its downward acceleration is :

         { 9 - 0.9t      if   0≤ x ≤ 10

a  =   { 0               if   x < 10

      
If the raindrop is initially m above the ground, how long does it take to fall?

 

I get that I need to take the antiderivatives of both parts (is there a more math-y word for that?) of the parametric function. The first antiderivative gives us the velocity and the second (again, right terminology? this is the first chapter on antiderivatives) gives us the position.

 

However, this is the function I got for s(t):

 

500 + 4.5t - .15 t³

 

So I plugged in 10 to see where the raindrop is after 10 seconds and I got 800, which is higher than where it started. What am I doing wrong? Also, how do I find the antiderivative of 0?

 

#73, p349

ANSWER: 11.8 secs

Answer 1

Your instinct is right — antidifferentiate twice — and two small things are throwing the answer off. Let me take your questions in order, then run the problem.

The terminology

The process is antidifferentiating, and the result is an antiderivative (or the indefinite integral). You already had the right word. One small correction: this is a piecewise-defined function, not a parametric one — parametric means x and y are each given in terms of a third variable.

The antiderivative of 0

It is C, an arbitrary constant. That is not a technicality here, it is the physics: once a = 0, velocity stops changing, so v is constant on that stretch. You pin down which constant by requiring v to be continuous at t = 10 — the raindrop does not teleport to a new speed.

What went wrong in your s(t)

You got s(t) = 500 + 4.5t² − 0.15t³, and two things slipped:

1. The initial velocity vanished. When you antidifferentiate a(t) to get v(t), the constant of integration is v(0) = 10, so v(t) = 10 + 9t − 0.45t². That 10 becomes a 10t term in the position function, and it is missing from yours.

2. The sign is backwards, and this is what produces the symptom you noticed. Antidifferentiating a downward velocity gives you distance fallen, which grows. Height above the ground shrinks. So you cannot add them — you need h(t) = 500 − s(t). Adding is exactly why your raindrop ended up higher than it started.

Pick one convention and stay in it. The cleanest here: let down be positive and let s(t) be distance fallen, with s(0) = 0. Then just ask when s(t) = 500.

Working it through

On 0 ≤ t ≤ 10, a(t) = 9 − 0.9t.

v(t) = 9t − 0.45t² + C, and v(0) = 10, so v(t) = 10 + 9t − 0.45t².

s(t) = 10t + 4.5t² − 0.15t³ + C, and s(0) = 0, so s(t) = 10t + 4.5t² − 0.15t³.

Now check where it is at t = 10, because that is where the rule changes:

s(10) = 100 + 450 − 150 = 400 m fallen — so it is still 100 m up, and it has not landed yet. Good, the second piece matters.

v(10) = 10 + 90 − 45 = 55 m/s.

For t > 10, a = 0, so v stays at 55 m/s (that is the constant C from earlier, fixed by continuity). The last 100 m at a steady 55 m/s takes 100/55 ≈ 1.82 s.

Total: 10 + 1.82 ≈ 11.8 seconds. That matches the book.

The habit to take from this

Every time you antidifferentiate in a motion problem, immediately ask what the constant means physically — v(0) for the first one, initial position for the second. Constants of integration are not bookkeeping in these problems; they are the initial conditions, and dropping one silently changes the situation you are modeling. The sign check is the same discipline: decide once whether up or down is positive, write it at the top of your page, and hold to it all the way through.