Corrected exercises: motion in a straight line (PHYS 101)

The graph is one straight line, so the slope never changes and the acceleration is constant. Read two points on the line, as far apart as possible to keep the reading error small: $(0 ; +8.0)$ and $(6.0 ; -4.0)$. Then $a = \dfrac{\Delta v}{\Delta t} = \dfrac{-4.0 - 8.0}{6.0 - 0} = \dfrac{-12.0}{6.0} = -2.0$ m/s². The minus sign is not an accident of the reading: the line goes down, so every second the velocity loses $2.0$ m/s, whatever the cart happens to be doing.

On a velocity-time graph the area between the line and the time axis is the displacement, counted with its sign. From $t = 0$ to $t = 4.0$ s the region is a triangle above the axis, base $4.0$ s and height $8.0$ m/s: $\Delta x_{1} = \tfrac{1}{2}(4.0)(8.0) = +16$ m. From $t = 4.0$ s to $t = 6.0$ s the region is a triangle below the axis, base $2.0$ s and height $4.0$ m/s: its area measures $\tfrac{1}{2}(2.0)(4.0) = 4.0$ m, and because the velocity is negative there the displacement is $\Delta x_{2} = -4.0$ m. Below the axis the cart is going backwards, so it gives back part of the ground it gained.

Total displacement: $\Delta x = +16 - 4.0 = +12$ m. It is a single arrow from the start point to the end point, and it is positive, so the cart finishes $12$ m to the right of where it began. Total distance travelled: add the areas without their signs, $16 + 4.0 = 20$ m. The two numbers differ because the cart turned around at $t = 4.0$ s. They would have been equal only if the velocity had kept one sign.

Average velocity uses the displacement: $v_{av} = \dfrac{\Delta x}{\Delta t} = \dfrac{+12}{6.0} = +2.0$ m/s. Average speed uses the distance: $\dfrac{20}{6.0} = 3.3$ m/s. Neither of them is the average of $+8.0$ and $-4.0$, which happens to give $+2.0$ here only because the acceleration is constant over the whole interval. Note the check that costs nothing: the average velocity must lie between the smallest and the largest velocity reached, and $+2.0$ does.

From $0$ to $4.0$ s the velocity is positive and the acceleration negative: opposite signs, so the speed falls, from $8.0$ m/s down to zero. The cart is moving to the right and slowing down. From $4.0$ s to $6.0$ s the velocity is negative and the acceleration is still negative: same signs now, so the speed grows, from zero to $4.0$ m/s. The cart is moving to the left and speeding up. The acceleration never changed, and it never needed to: its sign tells you which way the velocity is being pushed along your axis, not whether the object is gaining or losing speed. Writing the sentence the other way round, negative acceleration means slowing down, is the single most expensive habit of this chapter and it costs the whole of part e).

The displacement is the single arrow from the start point to the end point: $\Delta x = x_{f} - x_{i} = 10 - 0 = +10$ m. Nothing that happened in between enters this calculation, which is exactly what makes displacement cheap to compute and easy to misread. The distance travelled adds up the ground covered, leg by leg, always as a positive number: $30 + 20 = 50$ m. Five times the displacement, for the same walk.

Average velocity is displacement over elapsed time: $v_{av} = \dfrac{+10}{50} = +0.20$ m/s. Average speed is distance over the same elapsed time: $\dfrac{50}{50} = 1.0$ m/s. Note that the $10$ s standing still belongs in both denominators: the elapsed time is the time between the two instants named in the question, not the time spent moving. Dropping it would give $1.25$ m/s and $0.25$ m/s, and the error is invisible on the copy.

On the return leg she goes from $x = +30$ m to $x = +10$ m in $20$ s, so $v_{av} = \dfrac{10 - 30}{20} = \dfrac{-20}{20} = -1.0$ m/s. The minus sign is not a mistake and it is not optional: it says she is moving in the negative direction of the axis chosen in the statement, that is, back towards the door. If you had pointed the axis the other way at the start, the same walk would have given $+1.0$ m/s. The number depends on your axis, the physics does not, which is why the axis has to be written down before the first calculation.

A speed is a distance divided by a time, and both are positive, so a speed can never be negative. What the classmate computed is the average velocity, which is allowed to be negative. The correct pair of sentences is: her average velocity on the return leg is $-1.0$ m/s, her average speed on that leg is $1.0$ m/s. In an exam this confusion rarely costs one mark only, because the wrong sign is then carried into the next part.

Walking back the full $30$ m at $1.0$ m/s takes $30$ s, so the trip lasts $20 + 10 + 30 = 60$ s. She ends where she started: $\Delta x = 0$, hence $v_{av} = \dfrac{0}{60} = 0$ m/s exactly. Her average speed, on the other hand, is $\dfrac{30 + 30}{60} = 1.0$ m/s. A whole minute of walking with an average velocity of zero is not a paradox, it is the definition: the average velocity measures the net result of the trip, the average speed measures the effort. Any closed trip, on any path, has zero average velocity.

Read the curve: at $t = 1.0$ s the glider is at $x = 2.5$ m, at $t = 5.0$ s it is at $x = 14.5$ m. Average velocity is displacement over elapsed time, so $v_{av} = \dfrac{14.5 - 2.5}{5.0 - 1.0} = \dfrac{12.0}{4.0} = +3.0$ m/s. Geometrically this is the slope of the straight line joining the two points of the curve, the chord, and nothing in this calculation says anything about what happened between the two instants.

The instantaneous velocity at an instant is the slope of the TANGENT at that point, and the tangent is already drawn. Pick two points far apart on the dashed line so that the reading error stays small: it passes through $(1.0 ; 0.5)$ and $(5.0 ; 12.5)$. Slope $= \dfrac{12.5 - 0.5}{5.0 - 1.0} = \dfrac{12.0}{4.0} = +3.0$ m/s. Read the two points on the TANGENT, never one on the tangent and one on the curve: that is the classic way of losing this mark.

The two numbers are equal, $+3.0$ m/s, and this is not a coincidence. When the acceleration is constant the velocity grows linearly in time, so its average over an interval is exactly its value at the middle of that interval. Here $t = 3.0$ s is the midpoint of $[1.0 ; 5.0]$, so the chord and the tangent at the midpoint have the same slope. Rely on it only when the acceleration really is constant, that is, when the position-time graph is a parabola or a straight line. On any other curve, the chord and the tangent at the midpoint differ.

Use the same property backwards, choosing an interval centred on $t = 5.0$ s. At $t = 4.0$ s the curve gives $x = 10$ m, at $t = 6.0$ s it gives $x = 20$ m, so the average velocity over $[4.0 ; 6.0]$ is $\dfrac{20 - 10}{2.0} = +5.0$ m/s, and that is the instantaneous velocity at the midpoint $t = 5.0$ s. This trick replaces the tangent whenever the graph is a parabola and no tangent is drawn for you.

Two instantaneous velocities are now known, $+3.0$ m/s at $t = 3.0$ s and $+5.0$ m/s at $t = 5.0$ s, so $a = \dfrac{5.0 - 3.0}{5.0 - 3.0} = +1.0$ m/s², constant as announced. As for the last question: a velocity is a ratio of a change in position to a change in time, so it needs two readings, or a tangent, which is the limiting shape of a chord between two readings. A student who computes $\dfrac{x}{t} = \dfrac{6.5}{3.0} = 2.2$ m/s at $t = 3.0$ s has divided a position by a time, which is meaningful only when the motion started at the origin at $t = 0$ and never changed speed. Here it did both, and the answer is wrong.

Case 1: $v = +8$ and $a = +2$, both positive. The object moves in the positive direction and its velocity grows, so it speeds up. Case 2: $v = +8$ and $a = -2$. It still moves in the positive direction, but its velocity is being pulled down, so it slows down. Case 3: $v = -8$ and $a = -2$, both negative. It moves in the negative direction and its velocity becomes more negative, so its speed grows: it speeds up backwards. Case 4: $v = -8$ and $a = +2$. It moves backwards while its velocity is pulled up towards zero, so it slows down. Two of the four cases have a negative acceleration, and only one of those two is a slowing down.

The rule, in one sentence: if $v$ and $a$ have the SAME sign the speed increases, and if they have OPPOSITE signs the speed decreases. Nothing else is needed, and in particular the direction of travel is read off $v$ alone.

Average acceleration is the change in velocity over the elapsed time: $a = \dfrac{\Delta v}{\Delta t} = \dfrac{-12 - (-20)}{4.0} = \dfrac{+8.0}{4.0} = +2.0$ m/s². The acceleration is POSITIVE and yet the car is slowing down, because its speed went from $20$ m/s to $12$ m/s. This is case 4 of the figure. The brackets around $-20$ are not decoration: writing $-12 - 20$ instead gives $-8.0$ m/s², the exact opposite answer, and it is the most common arithmetic slip of the whole chapter.

Same calculation, the other way round: $a = \dfrac{-20 - (-12)}{4.0} = \dfrac{-8.0}{4.0} = -2.0$ m/s². The acceleration is negative and the car is speeding up, from $12$ m/s to $20$ m/s, still travelling backwards. This is case 3. Compare with c): the two motions have opposite accelerations and the same direction of travel, which is only possible because the sign of $a$ and the direction of travel are two independent pieces of information.

The sentence is false because it mixes up two different questions. The sign of $v$ answers which way the object is going along the axis you chose. The sign of $a$ answers which way the velocity is being changed along that same axis: positive means the velocity is being pushed towards the positive end of the axis, whether the object is going forwards, backwards or momentarily at rest. Only the comparison of the two signs says whether the speed is growing. The word deceleration hides this and is worth avoiding on a copy: write slowing down, and write the sign of $a$ with the axis you declared at the top of the page. A sentence such as I take the direction of the initial motion as positive, therefore $a < 0$ means slowing down here is correct and costs one line.

The statement gives $v_{i} = 0$, $a = +0.80$ m/s² and $\Delta x = +250$ m, and asks for $v_{f}$. No time appears anywhere, so take the equation without $\Delta t$: $v_{f}^{2} = v_{i}^{2} + 2a\,\Delta x = 0 + 2(0.80)(250) = 400$, hence $v_{f} = 20$ m/s. Keep the positive root: the train is moving in the positive direction. Check the order of magnitude, $20$ m/s is $72$ km/h, which is what a metro does between two distant stations.

Here $v_{i} = +25$ m/s, $v_{f} = 0$ and $\Delta t = 4.0$ s, and the question asks for $\Delta x$. The acceleration is neither given nor asked, so take the equation without $a$: $\Delta x = \dfrac{v_{i} + v_{f}}{2}\,\Delta t = \dfrac{25 + 0}{2}(4.0) = 50$ m. One line, no intermediate value, nothing to carry. A quick check: at constant $25$ m/s the cyclist would have covered $100$ m, and braking uniformly to rest halves that, which is exactly $50$ m.

Up the track is positive. Now $v_{i} = +6.0$ m/s, $a = -1.5$ m/s² and $\Delta t = 8.0$ s, with $v_{f}$ unknown and not asked: take the equation without $v_{f}$. $\Delta x = v_{i}\Delta t + \tfrac{1}{2}a\Delta t^{2} = (6.0)(8.0) + \tfrac{1}{2}(-1.5)(64) = 48 - 48 = 0$ m. The displacement is zero, which is a real answer and not an error: the cart stops after $\dfrac{6.0}{1.5} = 4.0$ s, having climbed $\dfrac{6.0 + 0}{2}(4.0) = 12$ m, then glides back down over the same $12$ m in the next $4.0$ s. The distance travelled is $12 + 12 = 24$ m. The return is real because its cause survives the stop: the slope pulls the cart down the track whether it is moving or not, so the acceleration keeps the value $-1.5$ m/s² before and after $t = 4.0$ s, and one equation covers the whole $8.0$ s. A puck slowed by friction on flat ice would behave differently: friction can only stop a motion, so the puck would stay at $12$ m after $t = 4.0$ s, and the equation would no longer apply past that instant. Answering $0$ m to the distance question, or $24$ m to the displacement question, are the two ways of losing this part.

The time is known and the displacement is now known too, but the shortest route uses the equation without $\Delta x$: $v_{f} = v_{i} + a\,\Delta t = 6.0 + (-1.5)(8.0) = 6.0 - 12.0 = -6.0$ m/s. The cart is moving at $6.0$ m/s in the NEGATIVE direction, that is, back down the track towards its starting point, which is consistent with a zero displacement: it passes the hand again at that very instant. Same speed as at the start, opposite velocity: this is the symmetry every uniformly accelerated motion has about the instant when it turns around.

Starting from $v_{f} = v_{i} + a\,\Delta t$ would give $a = \dfrac{0 - 25}{4.0} = -6.25$ m/s², and only then $\Delta x = (25)(4.0) + \tfrac{1}{2}(-6.25)(16) = 100 - 50 = 50$ m. The answer is the same, and so it should be, but the route is twice as long: one more line to write, one more number to carry, and above all one more sign to get right. In an exam under time pressure, the cost of that extra step is not the thirty seconds, it is the chance that the $-6.25$ becomes $+6.25$ on the next line and turns a braking into an acceleration. Choose the equation by the quantity that is MISSING, not by the one you remember best.

At the highest point the ball is neither rising nor falling, so $v = 0$ there. That is the physical translation of the word highest, and it is the only extra information the statement gives you. Take the equation without $\Delta x$: $0 = v_{i} + a\,\Delta t$, so $\Delta t = \dfrac{-v_{i}}{a} = \dfrac{-19.6}{-9.8} = 2.0$ s. Both minus signs are needed, and they cancel: a positive time, as it must be.

The time is now known, but the equation without $\Delta t$ is shorter and does not depend on the answer to a), so an error there cannot spread: $v_{f}^{2} = v_{i}^{2} + 2a\,\Delta x$ gives $0 = (19.6)^{2} + 2(-9.8)\Delta x$, hence $\Delta x = \dfrac{(19.6)^{2}}{2(9.8)} = \dfrac{384.16}{19.6} = 19.6$ m. The ball rises $19.6$ m above the hand. The height and the initial speed happen to share the same digits here, a coincidence of the numbers chosen, not a rule.

When the ball passes the hand again its displacement from the origin is zero, so $v_{f}^{2} = (19.6)^{2} + 2(-9.8)(0) = (19.6)^{2}$ and $v_{f} = \pm 19.6$ m/s. Choose the root that matches the physics: the ball is coming DOWN, so $v_{f} = -19.6$ m/s. Same speed as at launch, opposite direction. For the time, $v_{f} = v_{i} + a\Delta t$ gives $-19.6 = 19.6 - 9.8\,\Delta t$, so $\Delta t = \dfrac{39.2}{9.8} = 4.0$ s, exactly twice the time to the top. Reporting $+19.6$ m/s here is the classic loss: the number is a speed, the question asked for a velocity.

The catch happens $1.5$ m below the origin, so $\Delta x = -1.5$ m. Use the equation without $v_{f}$: $-1.5 = 19.6\,\Delta t - 4.9\,\Delta t^{2}$, that is $4.9\,\Delta t^{2} - 19.6\,\Delta t - 1.5 = 0$. The discriminant is $(19.6)^{2} + 4(4.9)(1.5) = 384.16 + 29.4 = 413.56$, whose square root is $20.336$. The two roots are $\Delta t = \dfrac{19.6 \pm 20.336}{9.8}$, that is $4.08$ s and $-0.075$ s. Reject the negative root: it describes where the ball would have come from had it been in free fall before the throw, which is not this problem. Velocity at the catch: $v_{f}^{2} = (19.6)^{2} + 2(-9.8)(-1.5) = 413.56$, so $v_{f} = -20.3$ m/s, slightly faster than at launch because the catch is lower than the throw.

The student is wrong. The velocity is zero at that instant, but the velocity is still CHANGING at that instant, from positive just before to negative just after, and it is exactly that change which the acceleration measures. Its value is $-9.8$ m/s² at the top as everywhere else on the flight, which is why the ball does not hang in the air. Distance and displacement for the flight of d): the ball goes up $19.6$ m and then down $19.6 + 1.5 = 21.1$ m, so the distance travelled is $40.7$ m, while the displacement is a single arrow from the hand to the catch, $-1.5$ m. One number is twenty-seven times the other for the same flight, and the only thing that tells the two questions apart is one word in the statement.

Upward positive, origin at the hand: the landing point is at $\Delta y = -h$, with $h > 0$.

The equation without $v_{f}$: $-h = v_{0}t - \tfrac{1}{2}gt^{2}$, that is $\tfrac{1}{2}gt^{2} - v_{0}t - h = 0$.

The square root is larger than $v_{0}$, so the root with the minus sign is negative: reject it, it lies before the throw.

Dimensions: $v_{0}^{2}$ is in m²/s² and so is $gh$, in (m/s²)(m), so the square root is in m/s like $v_{0}$, and m/s divided by m/s² gives seconds. A time, as it must be.

Limit case: as $h \to 0$, $t \to \dfrac{v_{0} + v_{0}}{g} = \dfrac{2v_{0}}{g} = \dfrac{2(19.6)}{9.8} = 4.0$ s, exactly the return time of part c).

Sense of variation: a larger $h$ makes the square root larger, so $t$ grows. A ball that has further to fall stays longer in the air, as it should.

$t = \dfrac{19.6 + \sqrt{(19.6)^{2} + 2(9.8)(8.0)}}{9.8} = \dfrac{19.6 + \sqrt{540.96}}{9.8} = \dfrac{19.6 + 23.26}{9.8} = 4.37$ s

With $h = 1.5$ m the same expression gives back the $4.08$ s of part d), a third check that costs one line.

Pitfall: writing $+h$ for a point below the hand gives $\sqrt{v_{0}^{2} - 2gh}$, a time that SHRINKS as the ball falls further. The sense-of-variation check catches that sign before any number is computed.

On an acceleration-time graph the area between the line and the time axis is the CHANGE of velocity, counted with its sign. Phase 1, from $0$ to $3.0$ s: a rectangle of height $2.0$ m/s² and width $3.0$ s, so $\Delta v = +6.0$ m/s. The cart started from rest, so $v(3.0) = 0 + 6.0 = 6.0$ m/s. Phase 2, from $3.0$ s to $7.0$ s: the acceleration is zero, the area is zero, the velocity does not change and $v(7.0) = 6.0$ m/s. Phase 3, from $7.0$ s to $9.0$ s: a rectangle of height $-3.0$ m/s² and width $2.0$ s, so $\Delta v = -6.0$ m/s and $v(9.0) = 6.0 - 6.0 = 0$ m/s. The cart is back at rest.

The velocity graph is now known: a straight line from $(0 ; 0)$ to $(3.0 ; 6.0)$, a horizontal segment at $6.0$ m/s until $t = 7.0$ s, then a straight line down to $(9.0 ; 0)$. Cut the area underneath into the two triangles and the rectangle drawn on the solution figure. Phase 1: $\tfrac{1}{2}(3.0)(6.0) = 9.0$ m. Phase 2: $(4.0)(6.0) = 24$ m. Phase 3: $\tfrac{1}{2}(2.0)(6.0) = 6.0$ m. Total displacement $9.0 + 24 + 6.0 = 39$ m. Everything is above the axis, so the distance travelled is also $39$ m.

Average velocity is the total displacement over the total time: $v_{av} = \dfrac{39}{9.0} = 4.3$ m/s. The formula $\dfrac{v_{i} + v_{f}}{2}$ would give $\dfrac{0 + 0}{2} = 0$, which is absurd since the cart clearly moved $39$ m. That formula is not a definition, it is a consequence of the velocity growing LINEARLY, which is true only while the acceleration stays constant. Over the whole $9.0$ s the acceleration takes three different values, so the formula does not apply. It would have been perfectly legitimate inside phase 1 alone, where $\dfrac{0 + 6.0}{2}(3.0) = 9.0$ m reproduces the area of the triangle.

No. A negative acceleration only says that the velocity is decreasing, and the velocity graph shows it falling from $6.0$ m/s to zero without ever crossing the axis. The cart keeps moving forwards, more and more slowly, and reaches rest exactly at $t = 9.0$ s. It would move backwards only if the negative acceleration lasted longer, pushing the velocity below zero. Here the recording stops at the instant the velocity reaches zero, and not one instant later.

The position-time graph has one shape per phase, and each follows from the velocity, since the slope of the position graph IS the velocity. Phase 1: the velocity grows from zero, so the graph starts flat and curves upwards, a piece of parabola, reaching $x = 9.0$ m at $t = 3.0$ s. Phase 2: constant velocity, so a straight line of slope $6.0$ m/s, from $9.0$ m to $33$ m. Phase 3: the velocity falls to zero, so the graph keeps rising but flattens out, a piece of parabola ending horizontal at $x = 39$ m. A curve that turns downwards, going back towards smaller positions, would contradict part d) and is the usual wrong sketch.

Upward positive, origin on the ground, $t = 0$ at take-off. The motion has two phases with one constant acceleration each, so the equations are used inside a phase, never across the fault.

Motors on, $a_{1} = +2.0$ m/s² for $4.0$ s:

$v_{1} = 0 + (2.0)(4.0) = 8.0$ m/s and $y_{1} = \tfrac{1}{2}(2.0)(4.0)^{2} = 16$ m

Motors off, $a_{2} = -9.81$ m/s². The fault changes the acceleration at once, but not the velocity: the second phase starts at $16$ m with $+8.0$ m/s.

Pitfall: writing $v = 0$ at the fault, as if the drone stopped there and dropped. It is still climbing at $8.0$ m/s.

Highest point, where $v = 0$: it comes $\dfrac{8.0}{9.81} = 0.82$ s after the fault, at $t = 4.82$ s, and the drone has climbed another $\dfrac{(8.0)^{2}}{2(9.81)} = 3.3$ m, to $16 + 3.3 = 19.3$ m.

Back at the altitude of the fault, the displacement of the second phase is zero:

$0 = 8.0\,t_{2} - 4.905\,t_{2}^{2}$, so $t_{2} = 0$ (the fault itself) or $t_{2} = \dfrac{8.0}{4.905} = 1.63$ s

The motors restart at $t = 4.0 + 1.63 = 5.63$ s, with $v = 8.0 - 9.81(1.63) = -8.0$ m/s: the same speed as at the fault, now downwards.

On the graph (figure below), after the fault the velocity follows one straight line of slope $-9.81$ m/s², from $+8.0$ m/s down through zero at $4.82$ s.

Between $4.0$ s and $4.82$ s it encloses a triangle above the axis, $\tfrac{1}{2}(0.82)(8.0) = +3.3$ m of climb; between $4.82$ s and $5.63$ s a triangle below the axis, $-3.3$ m of descent.

The two triangles are mirror images about the top, so their signed areas cancel: zero displacement since the fault, which is exactly the statement that the drone is back at $16$ m. It also shows why the climb and the descent each last $0.82$ s.

FALSE. A negative acceleration slows down an object that is moving in the positive direction, and speeds up one that is moving in the negative direction. Corrected version: a negative acceleration means the object is slowing down WHEN its velocity is positive, and speeding up when its velocity is negative. The compact form worth memorising: same signs for $v$ and $a$, the speed grows; opposite signs, the speed falls.

FALSE, and this is the statement that decides the mark on every throwing problem. A ball thrown upwards has exactly zero velocity at its highest point, and its acceleration there is $-9.8$ m/s², unchanged. If the acceleration really were zero at that instant, the velocity would have no reason to become negative and the ball would stay up there. Corrected version: a velocity of zero at one instant says nothing about the acceleration at that instant, because the acceleration measures how the velocity is CHANGING, not how big it is.

TRUE. On a velocity-time graph the area of a region above the time axis counts as a positive displacement and the area of a region below counts as a negative one, and adding them with their signs gives the displacement. Worth adding, because the next question in an exam usually asks for it: to obtain the DISTANCE travelled you add the same areas without their signs. The two agree only when the velocity keeps one sign over the whole interval.

FALSE, and the mistake is expensive because it survives into the whole calculation. In free fall the acceleration always points DOWNWARDS and its magnitude is $9.8$ m/s². Its sign is a consequence of the axis you declared: with the upward direction positive it is $-9.8$ m/s², with the downward direction positive it is $+9.8$ m/s². Corrected version: in free fall the acceleration is $9.8$ m/s² directed downwards, which is written $-9.8$ m/s² when the axis points up. A student who writes $-9.8$ while measuring depths positively downwards will find a negative depth and often keeps it.

TRUE, and it is the definition, not a special case. Note what it does NOT say: it does not say that the average velocity is the average of the initial and the final velocity, which is true only while the acceleration stays constant, and it does not say that it is the average of the speeds of the successive legs of a trip, which is almost never true. When in doubt, go back to this statement: total displacement, total elapsed time, divide.

Graph C.

From $0$ to $8.0$ s the acceleration keeps one value, $-1.5$ m/s², so the position graph is ONE parabola, curving downwards because $a < 0$.

Its slope is the velocity: steep at the start ($+6.0$ m/s), zero at the top ($t = 4.0$ s, $x = 12$ m), negative after, and it comes back to $x = 0$ at $t = 8.0$ s, symmetric about the top.

Graph A is two straight lines: a constant velocity up, then a constant velocity down, with a corner where the velocity would jump from $+6.0$ to $-6.0$ m/s in no time. Straight pieces mean zero acceleration, which is not this motion.

Graph B flattens and stays at $12$ m: that is a puck stopped by friction, which never comes back. On the tilted track the slope keeps pulling after the stop.

Graph D is a straight line falling through zero at $t = 4.0$ s: it is the VELOCITY graph, $v = 6.0 - 1.5t$, drawn on a position axis. Read as a position, it would put the cart above its launch point at $t = 0$ and below it at the end.

The five-second test: on a position-time graph, a horizontal tangent where $v = 0$, a straight piece where $a = 0$, and a curve bending downwards where $a < 0$. Only graph C passes all three.

With the origin and the instant $t = 0$ fixed by the statement, both motions start at $x = 0$. The cyclist leaves the line at rest and gains speed uniformly, so $x_{c} = v_{i}t + \tfrac{1}{2}at^{2} = 0 + \tfrac{1}{2}(1.2)t^{2} = 0.60t^{2}$. The runner keeps a constant velocity, so his acceleration is zero and $x_{r} = 6.0t$. Writing both positions from the SAME origin and the SAME clock is the whole difficulty of two-body problems; everything that follows is algebra.