Corrected exercises: kinematics in one and two dimensions (203-SN1-RE)

Differentiate: $v(t)=\frac{dx}{dt}=3t^{2}-12t+9$ (in m/s) and $a(t)=\frac{dv}{dt}=6t-12$ (in m/s²).

The cart is at rest when $v(t)=0$: $3t^{2}-12t+9=0$, that is $3(t-1)(t-3)=0$. So $t=1$ s and $t=3$ s.

$x(1)=1-6+9=4$ m and $x(3)=27-54+27=0$ m. The cart moves forward to 4 m, comes back to the origin, then sets off again.

Displacement: $x(4)-x(0)=(64-96+36)-0=4$ m, so the average velocity $=\frac{4}{4}=1$ m/s. Distance travelled: from 0 to 4 m (4 m), from 4 m back to 0 (4 m), then from 0 to 4 m (4 m), 12 m in all, so the average speed $=\frac{12}{4}=3$ m/s. The back-and-forth legs cancel in the displacement but add up in the distance: that is the whole difference between a vector and a scalar.

$a(1)=6-12=-6$ m/s² and $a(3)=18-12=6$ m/s². Yes, the cart reverses at both times: the velocity vanishes AND changes sign there (positive before $t=1$ s, negative between 1 s and 3 s, positive after 3 s). A zero velocity with a nonzero acceleration signals a turnaround, not a lasting stop.

The constant-acceleration equations assume that $a$ is constant, which is not the case here: $a$ decreases linearly. We must go back to the definitions and integrate. $v(t)=\int a\,dt=6t-0.75t^{2}+C_{1}$, and the shuttle starts from rest, so $v(0)=0$ gives $C_{1}=0$: $v(t)=6t-0.75t^{2}$ (m/s). Then $x(t)=\int v\,dt=3t^{2}-0.25t^{3}+C_{2}$, and $x(0)=0$ gives $C_{2}=0$: $x(t)=3t^{2}-0.25t^{3}$ (m).

$v(4)=6(4)-0.75(16)=24-12=12.0$ m/s. $x(4)=3(16)-0.25(64)=48-16=32.0$ m.

$a(t)=0$ when $6=1.50t$, that is $t=4.00$ s, exactly the instant the motor is cut. Since $a>0$ over the whole interval, the velocity has only increased: $v=12.0$ m/s is therefore the maximum velocity of the powered phase. A zero acceleration never means a zero velocity; it means a velocity that stops changing.

Average acceleration: $\bar{a}=\frac{v(4)-v(0)}{4}=\frac{12}{4}=3.00$ m/s$^{2}$. Average velocity: $\bar{v}=\frac{x(4)-x(0)}{4}=\frac{32}{4}=8.00$ m/s. Yet $\frac{v_{i}+v_{f}}{2}=\frac{0+12}{2}=6.00$ m/s, which is different. The formula $\bar{v}=\frac{v_{i}+v_{f}}{2}$ holds only for constant acceleration, because it assumes that the velocity grows linearly. Here the acceleration is strong at the start and weak at the end: the shuttle therefore spends more time at high velocity than at low velocity, and its average velocity exceeds the mean of the two extremes.

Second phase, constant acceleration this time, with $v_{i}=12.0$ m/s and $a=-0.500$ m/s$^{2}$. Duration: $\Delta t=\frac{12}{0.5}=24.0$ s, so the stop occurs at $t=4.00+24.0=28.0$ s. Extra distance: $\Delta x=\frac{v_{i}^{2}}{2|a|}=\frac{144}{1.00}=144$ m. Final position: $32.0+144=176$ m from the origin.

Phase 1 (0 to 8 s): $a_{1}=\frac{20-0}{8}=2.5$ m/s². Phase 2 (8 s to 28 s): constant velocity, $a_{2}=0$. Phase 3 (28 s to 36 s): $a_{3}=\frac{0-20}{8}=-2.5$ m/s² (braking).

The area under the curve is a trapezoid, which we cut up: triangle $\frac{1}{2}(8)(20)=80$ m, rectangle $(20)(20)=400$ m, triangle $\frac{1}{2}(8)(20)=80$ m. Total distance: $80+400+80=560$ m.

At $t=8$ s, the train has covered 80 m. Between 8 s and 10 s it travels at 20 m/s: $80+(20)(2)=120$ m.

$v_{avg}=\frac{560}{36}\approx 15.6$ m/s. It is below 20 m/s because the train spends part of the trip accelerating and braking: the average also weights these slower phases.

Three horizontal steps: $a=2.5$ m/s² from 0 to 8 s, $a=0$ from 8 s to 28 s, $a=-2.5$ m/s² from 28 s to 36 s, with jumps (discontinuities) at the phase changes. The acceleration is not defined continuously at the transition times, where the slope of the v-t graph changes abruptly.

With $y_{0}=19.6$ m and $v_{0}=+14.7$ m/s: $y(t)=19.6+14.7t-4.9t^{2}$ and $v(t)=14.7-9.8t$.

At the top, $v=0$: $t=\frac{14.7}{9.8}=1.5$ s. Height: $y(1.5)=19.6+\frac{(14.7)^{2}}{2(9.8)}=19.6+11.025=30.625$ m, about 30.6 m above the water.

Solve $y(t)=19.6$: $14.7t-4.9t^{2}=0$, that is $t(14.7-4.9t)=0$, so $t=3$ s (the other root is the launch). Velocity: $v(3)=14.7-29.4=-14.7$ m/s. Same magnitude as at launch, opposite sign: free fall is symmetric on either side of the top.

Solve $y(t)=0$: $19.6+14.7t-4.9t^{2}=0$. Dividing by $-4.9$: $t^{2}-3t-4=0$, that is $(t-4)(t+1)=0$. Keep $t=4$ s (the root $t=-1$ s precedes the throw and is rejected). Impact velocity: $v(4)=14.7-39.2=-24.5$ m/s, that is 24.5 m/s downward.

Average velocity: $\frac{\Delta y}{\Delta t}=\frac{0-19.6}{4}=-4.9$ m/s; the negative sign indicates that the net displacement is downward (the ball ends up below its starting point). Distance travelled: $11.025$ m on the way up then $30.625$ m on the way down, 41.65 m in all, hence an average speed of $\frac{41.65}{4}\approx 10.4$ m/s.

Uniform motion: $x_{c}(t)=25t$. Constant acceleration from rest: $x_{p}(t)=\frac{1}{2}(2.5)t^{2}=1.25t^{2}$.

The catch-up happens when $x_{p}=x_{c}$: $1.25t^{2}=25t$, that is $t(1.25t-25)=0$. Discard $t=0$ (the moment of passing): $t=20$ s, at $x=25(20)=500$ m.

$v_{p}(20)=(2.5)(20)=50$ m/s, that is 180 km/h. That is exactly double the velocity of the car being chased, and a very high velocity: in reality, a chase does not go on at constant acceleration to the very end.

The gap $e(t)=x_{c}(t)-x_{p}(t)=25t-1.25t^{2}$ is greatest when its derivative vanishes: $e'(t)=25-2.5t=0$, so $t=10$ s, that is when the two velocities are equal ($v_{p}=25$ m/s). As long as the police car is slower, the gap widens; as soon as it is faster, the gap closes. Greatest gap: $e(10)=250-125=125$ m.

General case: $x_{c}(t)=vt$ and $x_{p}(t)=\frac{1}{2}at^{2}$. The catch-up (other than $t=0$) satisfies $\frac{1}{2}at^{2}=vt$, so $t^{*}=\frac{2v}{a}$. The velocity of the pursuer is then $v_{p}=at^{*}=a\cdot\frac{2v}{a}=2v$. The acceleration drops out of the result: it decides only WHEN the catch-up happens, not the velocity at which it happens. Another way to see it: to catch a target that left at the same time, you need the same position, hence the same average velocity; starting from 0 with a velocity that grows linearly, the pursuer must finish at $2v$ to average $v$.

$\vec{r}(0)=0\hat{\imath}+1\hat{\jmath}$ and $\vec{r}(3)=\left(18-12\right)\hat{\imath}+10\hat{\jmath}=6\hat{\imath}+10\hat{\jmath}$ (m). Displacement: $\Delta\vec{r}=\vec{r}(3)-\vec{r}(0)=6\hat{\imath}+9\hat{\jmath}$ (m). Magnitude: $\left\|\Delta\vec{r}\right\|=\sqrt{6^{2}+9^{2}}=\sqrt{117}\approx 10.8$ m. Direction: $\tan\theta=\frac{9}{6}$, that is $\theta\approx 56.3^{\circ}$ north of east.

$\vec{v}_{avg}=\frac{\Delta\vec{r}}{\Delta t}=\frac{6\hat{\imath}+9\hat{\jmath}}{3}=2\hat{\imath}+3\hat{\jmath}$ (m/s), of magnitude $\sqrt{13}\approx 3.61$ m/s. Its direction is identical to that of the displacement, $56.3^{\circ}$ north of east: dividing a vector by a positive scalar changes its length, never its direction. The average velocity is always parallel to the displacement, however winding the actual path may be.

$\vec{v}(t)=\frac{d\vec{r}}{dt}=\left(4t-4\right)\hat{\imath}+3\hat{\jmath}$ (m/s) and $\vec{a}(t)=\frac{d\vec{v}}{dt}=4\hat{\imath}$ (m/s$^{2}$). The acceleration is constant in magnitude and direction: the motion is indeed uniformly accelerated. It is nevertheless not rectilinear, because $\vec{v}$ and $\vec{a}$ are not parallel: $\vec{a}$ points purely along $\hat{\imath}$ while $\vec{v}$ keeps a constant $\hat{\jmath}$ component. The path is a parabola, exactly like the path of a projectile turned on its side.

$\left\|\vec{v}\right\|^{2}=(4t-4)^{2}+9$. This sum is smallest when the square vanishes, that is $t=1.00$ s, and then $\left\|\vec{v}\right\|=3.00$ m/s. The drone does not stop: in two dimensions the velocity is zero only if BOTH components are, and $v_{y}=3$ m/s at all times. At $t=1$ s the drone merely stops moving east and heads due north.

At $t=3$ s: $\vec{v}=8\hat{\imath}+3\hat{\jmath}$, so $\left\|\vec{v}\right\|=\sqrt{73}\approx 8.54$ m/s, and $\vec{a}=4\hat{\imath}$ with $\left\|\vec{a}\right\|=4.00$ m/s$^{2}$. The tangential component is the projection of $\vec{a}$ onto the direction of $\vec{v}$: $a_{t}=\frac{\vec{a}\cdot\vec{v}}{\left\|\vec{v}\right\|}=\frac{4(8)+0(3)}{\sqrt{73}}=\frac{32}{\sqrt{73}}\approx 3.75$ m/s$^{2}$. The normal component makes up the rest of the magnitude: $a_{n}=\sqrt{\left\|\vec{a}\right\|^{2}-a_{t}^{2}}=\sqrt{16-14.03}\approx 1.40$ m/s$^{2}$. Interpretation: $a_{t}>0$ means the drone is speeding up, its speed is increasing; $a_{n}\neq 0$ means its path is curving. The turn can even be quantified by the radius of curvature $\rho=\frac{\left\|\vec{v}\right\|^{2}}{a_{n}}=\frac{73}{1.40}\approx 52.0$ m.

$v_{0x}=24.5\times 0.6=14.7$ m/s and $v_{0y}=24.5\times 0.8=19.6$ m/s. Hence $x(t)=14.7t$ (uniform motion) and $y(t)=19.6t-4.9t^{2}$ (constant acceleration).

At the top, $v_{y}=0$: $t=\frac{19.6}{9.8}=2$ s. $h_{max}=\frac{v_{0y}^{2}}{2g}=\frac{(19.6)^{2}}{19.6}=19.6$ m.

The flight ends when $y=0$: $t(19.6-4.9t)=0$, so $t=4$ s (twice the rising time, by symmetry). Range: $x(4)=14.7\times 4=58.8$ m.

The ball reaches the plane of the wall when $x=40$ m: $t=\frac{40}{14.7}\approx 2.72$ s. Height: $y(2.72)\approx 19.6(2.721)-4.9(2.721)^{2}\approx 53.33-36.28\approx 17.05$ m. Since 17.05 m $>$ 15 m, the ball clears the wall with a margin of about 2.1 m.

$v_{x}=14.7$ m/s (constant) and $v_{y}=19.6-9.8(2.721)\approx -7.07$ m/s. Magnitude: $\sqrt{(14.7)^{2}+(7.07)^{2}}\approx 16.3$ m/s. Direction: $\tan\alpha=\frac{7.07}{14.7}$, that is $\alpha\approx 25.7^{\circ}$ below the horizontal. The negative sign of $v_{y}$ shows that it is not: the ball has passed its top (reached at 2 s) and is already coming down as it crosses the wall.

$v_{0x}=25.0\cos 30^{\circ}=21.65$ m/s and $v_{0y}=25.0\sin 30^{\circ}=12.50$ m/s. Taking the ground as the origin of heights: $y(t)=20.0+12.50t-4.90t^{2}$. Impact occurs when $y=0$: $4.90t^{2}-12.50t-20.0=0$, so $t=\frac{12.50+\sqrt{156.25+392}}{9.80}=\frac{12.50+23.41}{9.80}\approx 3.66$ s. Keep the positive root; the other corresponds to a time before the throw. Range: $x=21.65\times 3.66\approx 79.3$ m.

$v_{x}=21.65$ m/s (unchanged) and $v_{y}=12.50-9.80(3.665)\approx -23.41$ m/s. Magnitude: $\sqrt{21.65^{2}+23.41^{2}}\approx 31.9$ m/s, directed $\arctan\frac{23.41}{21.65}\approx 47.2^{\circ}$ below the horizontal. General proof: $v_{x}=v_{0}\cos\theta$ does not change, and $v_{y}^{2}=v_{0}^{2}\sin^{2}\theta+2gh$ by the relation $v_{f}^{2}=v_{i}^{2}+2a\Delta y$ applied to the vertical motion. So $v^{2}=v_{0}^{2}\cos^{2}\theta+v_{0}^{2}\sin^{2}\theta+2gh=v_{0}^{2}+2gh$. Here $v=\sqrt{625+392}=\sqrt{1017}\approx 31.9$ m/s, whatever the angle. The angle decides how the velocity splits between $v_{x}$ and $v_{y}$, never the magnitude on arrival.

On level ground, the range is $R=\frac{v_{0}^{2}\sin 2\theta}{g}$. Setting $R=50.0$ m gives $\sin 2\theta=\frac{50.0\times 9.80}{625}=0.784$. The equation $\sin\alpha=0.784$ has two solutions in $[0^{\circ}, 180^{\circ}]$: $\alpha=51.6^{\circ}$ and $\alpha=180^{\circ}-51.6^{\circ}=128.4^{\circ}$. So $\theta_{1}\approx 25.8^{\circ}$ and $\theta_{2}\approx 64.2^{\circ}$. Their sum is $90^{\circ}$, which is the general result: $\sin 2\theta=\sin\left(180^{\circ}-2\theta\right)$, so $\theta$ and $90^{\circ}-\theta$ always give the same range. The flat shot and the lob are two faces of the same range.

Without calculation: the time of flight is $t=\frac{2v_{0}\sin\theta}{g}$, which depends only on the vertical component. The lob at $64.2^{\circ}$ sends much more velocity upward, so it stays in the air longer. Check: $t_{1}=\frac{2(25.0)\sin 25.8^{\circ}}{9.80}\approx 2.22$ s and $t_{2}=\frac{2(25.0)\sin 64.2^{\circ}}{9.80}\approx 4.59$ s, more than double. Same range, two very different flights: exactly the choice a tennis player or an artillery gunner has to make.