# Thread: A few antiderivative application problems

1. ## A few antiderivative application problems

Problem 1
A car starts from rest at time t = 0 and accelerates at -0.6t + 4 meters/sec2 for 0<t<12. How long does it take for the car to go 100 meters?

here is my work:
$a = -0.6t + 4$
$v = -0.3t^2 + 4t + c$
$v = -0.3t^2 + 4t$
$s = -0.1t^3 + 2t^2 + K$
$s = -0.1t^3 + 2t^2$
$100 = -0.1t^3 + 2t^2$
$t = 10 seconds$

Problem 2
A 727 jet needs to attain a speed of 200 mph to take off. If it can accelerate from 0 to 200 mph in 30 seconds, how long must the runway be? (Assume constant acceleration and give your final answer in feet)

here is my work:
200 miles per hour = $293 \frac{1}{3}$ ft/sec

$293 \frac{1}{3}$ ft/sec divided by 30 sec = $9 \frac{7}{9}$ ft/sec $^2$

$a = 9 \frac{7}{9}$ ft/sec $^2$
$v = 9 \frac{7}{9} t + c$ ft/sec
$v = 9 \frac{7}{9} t$ ft/sec
$s = 4 \frac{8}{9} t^2 + K$
$s = 4 \frac{8}{9} t^2$
$s(30) = 4 \frac{8}{9} * (30)^2$
$s(30) = 4400$ ft.

Problem 3
On the moon, the acceleration due to gravity is about 1.6 m/sec $^2$ (compared to $g \approx 9.8$ m/sec $^2$ on earth). If you drop a rock on the moon (with initial velocity 0), find formulas for:

(a) Its velocity, v(t), at time t.
(b) The distance, s(t), it falls in time t.

here is my work:
$a = -1.6$ m/sec $^2$
$v = -1.6t + c$ m/sec
(a) $v(t) = -1.6t$ m/sec
(b) $s(t) = -0.8t^2 + K$, where K is the initial height.

Should I keep the constant K in this answer? and how about the constants in all the other problems? I'm confused whether I need them or not.. In the previous ones I assumed the initial values were 0. Can I do this?

2. Originally Posted by Jacobpm64
Problem 1
A car starts from rest at time t = 0 and accelerates at -0.6t + 4 meters/sec2 for 0<t<12. How long does it take for the car to go 100 meters?
$a=-.6t+4$
$v=\int -.6t+4dt=.3t^2+4t+C$
At $t=0$ we have $v=0$ (meaning of "at rest").
Thus, $C=0$.
Thus,
$v=-.3t^2+4t$
$s=\int -.3t^2+4t dt$
$s=-.1t^3+2t^2+C$
Is the equation for the distance function.

3. All right,

But then how would you get a value for C?

4. Originally Posted by Jacobpm64
All right,

But then how would you get a value for C?
Look at what he wrote he
tell you that $C=0$, which leaves you with:

$s=-0.1\, t^3 +2\,t^2$

To find how long it takes to go 100 metres you set $s=100$ in the above equation and solve for $t$. Which gives you the
following cubic to solve:

$-0.1\, t^3 +2\,t^2-100=0$,

which the observant will note has a root at $t=10$, and the other roota are either negative ( and so irrelevant) or at a time greater than 12 (and so also irrelevant).

So it takes 10 seconds for the car to go 100 metres.

RonL