Corrected exercises: rectilinear motion (201-SN2-RE)

$v(t)=-6t+18$ and $a(t)=-6$. So $v(1)=12$ m/s and $a(1)=-6$ m/s$^{2}$: the particle moves forward and slows down, since $v$ and $a$ have opposite signs.

$v(t)=6t^{2}-30t+36=6(t^{2}-5t+6)=6(t-2)(t-3)$. The particle is at rest at $t=2$ s and at $t=3$ s. Factor out $6$ BEFORE looking for the roots: what remains is a trinomial with integer roots.

$v(t)=3t^{2}-6t+6=3\left[(t-1)^{2}+1\right]$, whose discriminant $36-72$ is negative. So $v(t)\ge 3>0$ for all $t$: the particle never stops. The minimum of the velocity is $v(1)=3$ m/s, at the time when $a(t)=6t-6$ is zero. Pitfall: answering $t=1$, which makes the acceleration zero, not the velocity.

$v(t)=4t^{3}-12t^{2}+8t=4t(t^{2}-3t+2)=4t(t-1)(t-2)$: at rest at $t=0$, $t=1$ and $t=2$, the last one at $t=2$ s. Then $a(t)=12t^{2}-24t+8$, so $a(2)=48-48+8=8$ m/s$^{2}$. Pitfall: dividing $v=0$ by $t$ and losing the root $t=0$; here it is not the answer, but it is one of the times at rest.

$s(t)=(2t+1)^{\frac{1}{2}}$, so $v(t)=\frac{1}{2}(2t+1)^{-\frac{1}{2}}\cdot 2=(2t+1)^{-\frac{1}{2}}$ and $a(t)=-\frac{1}{2}(2t+1)^{-\frac{3}{2}}\cdot 2=-(2t+1)^{-\frac{3}{2}}$. At $t=4$, $2t+1=9$: $v(4)=\frac{1}{3}$ m/s and $a(4)=-\frac{1}{27}$ m/s$^{2}$. Pitfall: forgetting the factor $2$ from the inner derivative, which gives $\frac{1}{6}$.

Quotient rule: $v(t)=\frac{10(t+2)-10t}{(t+2)^{2}}=\frac{20}{(t+2)^{2}}=20(t+2)^{-2}$, then $a(t)=-40(t+2)^{-3}$. At $t=2$: $v(2)=\frac{20}{16}=1.25$ m/s and $a(2)=-\frac{40}{64}=-0.625$ m/s$^{2}$. Once $v$ is simplified, the second derivative is done with the power rule, without going through the quotient rule again.

$v(t)=-9.8t+14.7$, zero at $t=\frac{14.7}{9.8}=1.5$ s. Then $s(t)=-4.9(t^{2}-3t-4)=-4.9(t-4)(t+1)$, zero at $t=4$ s for $t\ge 0$. So $v(4)=-39.2+14.7=-24.5$ m/s. Pitfall: giving $24.5$; the particle is moving down, its velocity is NEGATIVE.

$v(t)=6e^{-0.5t}\cdot(-0.5)+3=3-3e^{-0.5t}$ and $a(t)=-3e^{-0.5t}\cdot(-0.5)=1.5e^{-0.5t}$. At $t=2$: $v(2)=3-\frac{3}{e}\approx 1.896$ m/s and $a(2)=\frac{1.5}{e}\approx 0.552$ m/s$^{2}$. Pitfall: the factor $-0.5$ from the exponent, forgotten or taken without its sign.

$v(t)=1+2\cos t$, zero when $\cos t=-\frac{1}{2}$, that is $t=\frac{2\pi}{3}$ and $t=\frac{4\pi}{3}$ on $[0, 2\pi]$: the first time at rest is $t=\frac{2\pi}{3}\approx 2.094$ s. Then $a(t)=-2\sin t$, so $a\left(\frac{2\pi}{3}\right)=-2\cdot\frac{\sqrt{3}}{2}=-\sqrt{3}\approx-1.732$ m/s$^{2}$.

$v(t)=2t-\frac{6}{t+1}$. $v=0$ gives $2t(t+1)=6$, that is $t^{2}+t-3=0$, and the only positive root is $t=\frac{-1+\sqrt{13}}{2}\approx 1.303$ s. Then $a(t)=2+\frac{6}{(t+1)^{2}}$; since $t^{2}+t=3$, $(t+1)^{2}=t^{2}+2t+1=t+4$, so $a=2+\frac{6}{t+4}\approx 2+\frac{6}{5.3028}\approx 3.131$ m/s$^{2}$. Pitfall: the derivative of $\ln(t+1)$ is $\frac{1}{t+1}$, not $\frac{1}{t}$.

$v(t)=s'(t)=3t^{2}-18t+24$, in metres per second, and $a(t)=v'(t)=6t-18$, in metres per second squared. Factor the velocity, which will be useful everywhere afterward: $v(t)=3\left(t^{2}-6t+8\right)=3(t-2)(t-4)$. A factored velocity is a velocity whose sign you read without any calculation.

At $t=1$: $s(1)=1-9+24=16$ metres, $v(1)=3-18+24=9$ metres per second, $a(1)=6-18=-12$ metres per second squared. At $t=5$: $s(5)=125-225+120=20$ metres, $v(5)=75-90+24=9$ metres per second, $a(5)=30-18=12$ metres per second squared. The two times have the same velocity and opposite accelerations: at $t=1$ the particle slows down, at $t=5$ it speeds up.

They are the points where $v(t)=0$, that is $t=2$ and $t=4$. There we find $s(2)=8-36+48=20$ metres and $s(4)=64-144+96=16$ metres. Physically, these are the TURNAROUND times: the particle stops for an instant and sets off in the opposite direction. On the figure, the curve has a horizontal tangent there, a local maximum then a local minimum.

$a(t)=0$ gives $t=3$, and then $v(3)=27-54+24=-3$ metres per second. This is the most NEGATIVE velocity of the motion, so the moment when the particle moves backward fastest: the velocity, a quadratic function with its vertex at $t=3$, reaches its minimum there. That is the physical meaning of “zero acceleration”: no forward push, no braking, the velocity is at an extremum.

The claim confuses zero acceleration with zero velocity. At $t=3$ the velocity is $-3$ metres per second, so the particle is moving backward at 3 metres per second, and it has never been faster in that direction. Rest means $v=0$, which happens only at $t=2$ and $t=4$. The error is the exact mirror of believing that an extremum of $s$ is an extremum of $v$: each derivative speaks about the function before it, not about the others.

$v$ is zero at 2 and at 4, and its leading coefficient is positive: it is therefore positive outside the roots and negative between them. On $[0, 2)$: $v>0$, the particle moves forward, in the positive direction. On $(2, 4)$: $v<0$, it moves backward. On $(4, 5]$: $v>0$, it moves forward again. Check at three points: $v(1)=9>0$, $v(3)=-3<0$, $v(5)=9>0$ ✓.

$s$ is increasing exactly where $v=s'>0$, so on $[0, 2]$ and on $[4, 5]$, and decreasing on $[2, 4]$. This is not one more result: it is the same statement as part a, translated from the vocabulary of physics into that of mathematics. “The particle moves backward” and “the position decreases” are two ways of saying that $s'<0$.

Yes, and the turnaround is precisely what makes it possible. For example $s(1)=16$ and $s(4)=16$: the particle is at the same place at two different times. You can check it another way: $s(t)=16$ reads $t^{3}-9t^{2}+24t-16=0$, of which $t=1$ is an obvious root, and factoring gives $(t-1)(t^{2}-8t+16)=(t-1)(t-4)^{2}$. The solutions are therefore $t=1$ and $t=4$, the latter being a double root, which reflects the fact that the particle arrives there while stopping.

Solve $t^{3}-9t^{2}+24t=18$, that is $t^{3}-9t^{2}+24t-18=0$. The obvious root $t=3$ works: $27-81+72-18=0$ ✓. Factor: $(t-3)\left(t^{2}-6t+6\right)=0$, whose other roots are $t=3\pm\sqrt{3}$, that is $1.2679$ and $4.7321$. All three values lie in $[0, 5]$: the particle therefore passes THREE times through the position 18 metres. This is the direct reading of the graph, which crosses the same horizontal line three times.

A function has an inverse only if it is one-to-one, that is, if each value is taken only once. Here $s$ is not monotonic on $[0, 5]$, it takes the value 18 three times: it is not one-to-one, so not invertible. Practical consequence: the question “at what time is the particle at 18 metres” does not have ONE answer, it has three, and you must give them all. You only get an inverse back by restricting to an interval on which the function is monotonic, for example $[0, 2]$, where the answer is the single $t=1.2679$.

The displacement is $s(5)-s(0)=20-0=20$ metres. It is a difference of two positions, nothing more: the path followed in between plays no role. A displacement can even be zero although the particle has moved a lot, if it comes back to its starting point.

Split at the times where $v$ changes sign, that is $t=2$ and $t=4$. From 0 to 2: $\left|s(2)-s(0)\right|=\left|20-0\right|=20$ metres. From 2 to 4: $\left|16-20\right|=4$ metres. From 4 to 5: $\left|20-16\right|=4$ metres. Total: $20+4+4=28$ metres. The figure shows the three pieces, with the two short back-and-forth trips of 4 metres. Not splitting would give 20 metres, and the error is 8 metres, that is 40 percent.

Average velocity: $\frac{s(5)-s(0)}{5-0}=\frac{20}{5}=4$ metres per second. Average speed: $\frac{28}{5}=5.6$ metres per second. A car's speedometer shows the ABSOLUTE value of the instantaneous velocity, so accumulating it gives the second one: the distance shown by the odometer is the one from part b, never the displacement. No odometer runs backward when the car reverses.

Solve $v(c)=4$, that is $3c^{2}-18c+24=4$, so $3c^{2}-18c+20=0$ and $c=\frac{18\pm\sqrt{324-240}}{6}=\frac{18\pm\sqrt{84}}{6}$. The two roots are $1.4725$ and $4.5275$, both in $(0, 5)$: the theorem promised at least one, there are two, which is consistent with it being an existence statement.

They agree exactly when the velocity KEEPS A CONSTANT SIGN over the interval, that is, when the particle does not turn around. On $[0, 2]$ for example: displacement 20 metres, distance 20 metres, both averages equal 10 metres per second. This is the condition to check before skipping the splitting, and it fits in one line: is the velocity zero anywhere inside the interval?

The speed is $\left|v\right|=\sqrt{v^{2}}$, whose derivative is $\frac{2vv'}{2\sqrt{v^{2}}}=\frac{va}{\left|v\right|}$ for $v\neq 0$. Since $\left|v\right|>0$, the sign of this derivative is that of $va$. So $\left|v\right|$ increases if and only if $va>0$, that is, if $v$ and $a$ have the same sign. The reasoning is purely mathematical, but what it states is the physical principle: a force in the direction of motion speeds the object up, an opposing force slows it down.

$v$ is positive on $[0, 2)$, negative on $(2, 4)$, positive on $(4, 5]$. $a=6t-18$ is negative on $[0, 3)$ and positive on $(3, 5]$. Product: negative on $[0, 2)$, POSITIVE on $(2, 3)$, negative on $(3, 4)$, positive on $(4, 5]$.

The particle SPEEDS UP where the product is positive, so on $(2, 3)$ and on $(4, 5]$. It SLOWS DOWN on $[0, 2)$ and on $(3, 4)$. The result is surprising: between 2 and 3, the acceleration is negative and yet the particle goes faster and faster. That is because it is moving backward, and a negative acceleration pushes precisely in the direction it is going. No sign chart of $a$ alone could have given this answer.

At $t=1$: $v=9$ and $a=-12$, of opposite signs, so the particle SLOWS DOWN. Check: $v(1.5)=3(2.25)-27+24=6.75-3=3.75$ metres per second, against 9 at $t=1$. The speed has indeed gone from 9 to 3.75. Note that the velocity has decreased as well, but that is a coincidence of this case: between 2 and 3, the velocity decreases while the speed increases.

Yes, at $t=3$, where $a=0$ and $v=-3$ metres per second. The product $va$ is zero there and changes sign: the speed goes through a local MAXIMUM, equal to 3 metres per second. It is the fastest point of the backward phase. Physically, it is the time when the force changes direction: it stops pushing the particle backward and starts braking it.

$y(t)=-4.9t^{2}+20t+25$ metres, $v(t)=y'(t)=-9.8t+20$ metres per second, $a(t)=v'(t)=-9.8$ metres per second squared, constant. The coefficient $-4.9$ is $\frac{-g}{2}$, the minus sign reflecting the choice of an upward axis. It is this choice, not the physics, that sets the signs: always state it before computing.

The top corresponds to $v=0$, so $t=\frac{20}{9.8}\approx 2.0408$ seconds. The height is then $y(2.0408)=-4.9\times 4.1649+20\times 2.0408+25=-20.408+40.816+25=45.408$ metres. You can check it another way: the vertex of a parabola $at^{2}+bt+c$ is at $t=\frac{-b}{2a}=\frac{-20}{-9.8}$, the same value.

The ground corresponds to $y=0$, that is $4.9t^{2}-20t-25=0$, so $t=\frac{20+\sqrt{400+490}}{9.8}=\frac{20+\sqrt{890}}{9.8}\approx\frac{49.833}{9.8}\approx 5.0850$ seconds; the negative root is rejected, since it corresponds to a past in which the motion did not exist. The impact velocity is $v(5.0850)=-9.8\times 5.0850+20\approx-29.833$ metres per second, that is about 29.8 metres per second downward, or 107 kilometres per hour.

$v^{2}=20^{2}+2\times 9.8\times 25=400+490=890$, so $\left|v\right|=\sqrt{890}\approx 29.833$ ✓, exactly the value found. This relation is obtained by eliminating the time between $v=v_{0}-gt$ and $y=y_{0}+v_{0}t-\frac{g}{2}t^{2}$, which amounts to the work-energy theorem. Its advantage is that it answers without ever solving the quadratic equation in $t$: when the question is not about the time, you can skip computing it.

The starting height is 25 metres: solve $-4.9t^{2}+20t+25=25$, that is $t(-4.9t+20)=0$, so $t=0$, the throw, and $t=\frac{20}{4.9}\approx 4.0816$ seconds. The velocity there is $v(4.0816)=-9.8\times 4.0816+20=-20$ metres per second: exactly the OPPOSITE of the launch velocity. The stone passes the same point again at the same speed, a consequence of the symmetry of the parabola, and it is also conservation of mechanical energy read off a graph.

Phase 1, from 0 to 2 seconds: the velocity increases linearly from 0 to 3 metres per second, so the acceleration is constant and equals $\frac{3-0}{2-0}=+1.5$ metres per second squared. Phase 2, from 2 to 8 seconds: the velocity is constant at 3 metres per second, the acceleration is zero. Phase 3, from 8 to 10 seconds: the velocity decreases from 3 to 0, the acceleration is $\frac{0-3}{10-8}=-1.5$ metres per second squared.

The area under the velocity curve represents the DISPLACEMENT, here the height climbed. It is a trapezoid with bases of 10 and 6 seconds and a height of 3 metres per second: area $=\frac{10+6}{2}\times 3=24$ metres. You can also split it up: two triangles of $\frac{1}{2}\times 2\times 3=3$ metres each, plus a rectangle of $6\times 3=18$ metres, total $3+18+3=24$ metres ✓. The elevator climbs 24 metres, about eight floors.

The position starts at 0 and keeps increasing, since $v\geq 0$ everywhere. Phase 1: $v$ increases, so $s''=a>0$, the curve is CONCAVE UP. Phase 2: $a=0$, the position is a STRAIGHT LINE SEGMENT with slope 3. Phase 3: $a<0$, the curve is CONCAVE DOWN and flattens out until it becomes horizontal at $t=10$. The overall shape is a stretched-out S, what is called a ramp-shaped curve.

The graph of $a$ is made of three horizontal steps: $+1.5$ from 0 to 2, then 0 from 2 to 8, then $-1.5$ from 8 to 10. It is DISCONTINUOUS at $t=2$ and $t=8$ because the velocity has a CORNER there, an abrupt change of slope: at those points $v$ is not differentiable, so $a$ is not defined there and jumps. A real elevator smooths these transitions, precisely to avoid the jolt that this jump represents; the derivative of the acceleration is called the jerk.

The scale measures the normal force $N$, given by $N=m(g+a)$. Phase 1: $N=70\times(9.8+1.5)=791$ newtons, an apparent mass of $\frac{791}{9.8}\approx 80.7$ kilograms, about 15 percent more. Phase 2: $N=70\times 9.8=686$ newtons, the true weight, the passenger feels nothing. Phase 3: $N=70\times(9.8-1.5)=581$ newtons, that is $59.3$ apparent kilograms, the familiar feeling of lightness on arrival. The scale never measures mass, it measures a force, and the acceleration makes the difference.