Calculus BC ยท Unit 8
Integrate velocity for displacement. Integrate speed for distance, and split where the velocity changes sign.
Velocity accumulates into position
\(s(b)=s(a)+\int_a^b v(t)\,dt\)
The integral of velocity is displacement, the net change in position.
Distance uses the absolute value
Total distance is
\(\int_a^b |v(t)|\,dt\)
Split the integral where \(v\) changes sign.
Speed and direction
Speed is \(|v|\). The particle moves right when \(v>0\) and left when \(v<0\).
Acceleration is the derivative of velocity
\(a(t)=v'(t)\). Speeding up means velocity and acceleration have the same sign.
Velocity does not change sign.
\(\int_0^4 3\,dt\)
\(12\)
\(s(4)=1+12\)
\(13\)
Answer: displacement 12 ft, \(s(4)=13\)
Displacement: \(\int_0^4 -2\,dt=-8\).
Speed is 2 the whole time.
Distance: \(\int_0^4 2\,dt=8\).
\(s(4)=9-8=1\)
Answer: displacement \(-8\), distance 8
\(\int_0^3 4t\,dt=\left[2t^2\right]_0^3\)
\(18\)
\(s(3)=2+18=20\)
Answer: \(20\)
\(t+1>0\) for every \(t\ge 0\).
It is moving right on the whole interval.
Answer: for all \(t\ge 0\)
\(v(t)=3t^2+C\)
\(v(0)=C=-4\)
\(v(t)=3t^2-4\)
Answer: \(3t^2-4\)
\(\dfrac{12}{4}\)
\(3\)
Answer: \(3\)
Velocity changes sign. Split the distance integral.
Velocity is zero at \(t=2\).
\(\int_0^3 (2t-4)\,dt=\left[t^2-4t\right]_0^3\)
\(9-12\)
\(-3\)
Distance on \([0,2]\): \(\int_0^2 (4-2t)\,dt=\left[4t-t^2\right]_0^2=4\).
Distance on \([2,3]\): \(\left[t^2-4t\right]_2^3=1\).
Total distance: \(5\).
\(s(3)=5+(-3)=2\)
Answer: displacement \(-3\), distance 5, \(s(3)=2\)
\(v(t)<0\) on \([0,2)\), so it moves left there.
\(a(t)=2>0\) always.
Same sign: speeding up when \(v>0\), which is \((2,3]\).
Opposite signs on \([0,2)\): slowing down while moving left.
Answer: left on \([0,2)\); speeding up on \((2,3]\)
Displacement: \(\int_0^{\pi}\sin t\,dt=2\).
\(\sin t\ge 0\) on \([0,\pi]\), so distance equals displacement.
\(s(\pi)=2\)
Answer: both are 2
| t | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| v(t) | 4 | 1 | \(-2\) | \(-3\) |
\(\dfrac{2}{2}\left(4+2\cdot 1+2\cdot(-2)+(-3)\right)\)
\(4+2-4-3\)
\(-1\)
Answer: \(-1\) foot
Speeds: 4, 1, 2, and 3.
\(4+2\cdot 1+2\cdot 2+3\)
\(13\)
This approximates \(\int |v|\), not the signed integral.
Answer: 13 feet
\(v(t)=4t-6\)
\(s(t)=2t^2-6t+1\)
\(s(2)=8-12+1\)
\(-3\)
Answer: \(-3\)
A zero inside the interval, or a table that must not be treated as distance.
\(t^2-4=0\) at \(t=2\) in the interval. \(t=-2\) is outside.
On \([0,2]\), velocity is negative. Distance: \(\int_0^2 (4-t^2)\,dt\).
\(\left[4t-\dfrac{t^3}{3}\right]_0^2\)
\(8-\dfrac{8}{3}\)
\(\dfrac{16}{3}\)
On \([2,3]\): \(\int_2^3 (t^2-4)\,dt=\left[\dfrac{t^3}{3}-4t\right]_2^3\).
\((9-12)-\left(\dfrac{8}{3}-8\right)\)
\(-3-\left(-\dfrac{16}{3}\right)\)
\(\dfrac{7}{3}\)
Total: \(\dfrac{16}{3}+\dfrac{7}{3}=\dfrac{23}{3}\).
Answer: \(\dfrac{23}{3}\)
\(\int_0^3 (t^2-4)\,dt=\left[\dfrac{t^3}{3}-4t\right]_0^3\)
\(9-12\)
\(-3\)
The leftward part and the rightward part cancel in the signed integral.
Answer: \(-3\)
The integral of a function that changes from positive to negative can be negative.
The two given values do not determine the integral.
The claim is not justified.
Answer: no
It stops when \(6-2t=0\), so \(t=3\).
\(s(t)=6t-t^2+4\)
\(s(3)=18-9+4=13\)
\(s(5)=30-25+4=9\)
Answer: \(s(3)=13\) and \(s(5)=9\)
Velocity and acceleration are both negative.
Same sign means the speed is increasing.
The particle is speeding up while moving left.
Answer: speeding up
Distance is the leftward travel plus the rightward travel.
Net displacement \(-6\) means leftward travel exceeds rightward travel by 6.
Let rightward travel be \(r\) and leftward travel be \(\ell\).
\(\ell+r=10\) and \(r-\ell=-6\).
Add them: \(2r=4\), so \(r=2\) and \(\ell=8\).
Answer: 2 units right and 8 units left
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