Scalars and vectors in one dimension

The equations and values this unit uses that the exam's Table of Information prints: the three constant-acceleration equations $v_x = v_{x0} + a_xt$, $x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2$ and $v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$; the right-triangle relationships $a^2 + b^2 = c^2$, $\sin\theta = \frac{a}{c}$, $\cos\theta = \frac{b}{c}$ and $\tan\theta = \frac{a}{b}$; the sine, cosine and tangent of 0°, 30°, 37°, 45°, 53°, 60° and 90°; and the acceleration due to gravity at Earth's surface, printed as $g = 9.8\ \text{m/s}^2$.

• Displacement $\Delta x = x - x_0$, average velocity $\vec{v}_{avg} = \frac{\Delta\vec{x}}{\Delta t}$ and average acceleration $\vec{a}_{avg} = \frac{\Delta\vec{v}}{\Delta t}$.

• For constant acceleration, the average velocity $\frac{v_{x0} + v_x}{2}$, which is also the instantaneous velocity at the middle of the time interval.

• Slope and area on motion graphs: velocity is the slope of position against time, acceleration the slope of velocity against time, displacement the area under velocity against time, change in velocity the area under acceleration against time.

• The value $g \approx 10\ \text{m/s}^2$ used for numerical answers.

• Relative velocity along one line: the velocity of object A measured by observer B is $v_A - v_B$.

• Components of a vector of magnitude $v_0$ at angle $\theta$ above the horizontal: $v_0\cos\theta$ and $v_0\sin\theta$.

• Projectile results: time to the top $\frac{v_{y0}}{g}$, maximum height $\frac{v_{y0}^2}{2g}$, time of flight and range on level ground, and a launch speed from a maximum height. Each is derived from the constant-acceleration equations whenever it is needed.

• Section I has 42 multiple-choice questions in 85 minutes, each with four options and one correct answer. Section II has four free-response questions in 95 minutes, in a fixed order: Mathematical Routines (MR, 10 points), Translation Between Representations (TBR, 12 points), Experimental Design and Analysis (LAB, 10 points) and Qualitative/Quantitative Translation (QQT, 8 points). The question paper prints no point values.

• Every scoring element is one point, labeled by its part (Point A1, Point A2, Point B1 and so on).

• A derivation earns a point for a multistep derivation that starts from the named law or a table equation, one or two points for the right substitutions, and a point for the final expression. A correct final expression written alone earns all of these except the first, so the principle line is the one point a bare answer never earns.

• A comparison ("greater than, less than, or equal to") earns a point for the choice and a point for each physics link: what stays the same, what changes, and the relationship that connects them. Quoting an equation is not a justification.

• A consistency part earns a point for any attempt to use functional dependence ("proportional to", "in the denominator") and a second point for using it correctly against both your claim and your expression.

• A part earns credit "consistent with" an earlier answer only where the scoring says so. Extra or unlabeled arrows on a diagram cap that part. Numerical answers are credited inside a range, and every final number should carry its unit. "Human error" never earns a point.

A robot vacuum moves along a straight hallway.

Taking east as positive, it moves 3.0 m east, then 5.0 m west, then 1.5 m east.

Calculate its total displacement and the distance it travels.

Step 1: Write each displacement as a signed component: $+3.0$ m, $-5.0$ m, $+1.5$ m.

Step 2: Add the components: $\Delta x = (+3.0) + (-5.0) + (+1.5) = -0.5$ m.

Step 3: Add the magnitudes for the distance: $3.0 + 5.0 + 1.5 = 9.5$ m.

Answer: displacement 0.5 m west ($-0.5$ m); distance 9.5 m.

The negative sign is a direction, not a "negative amount" of motion.

A displacement can be much smaller than the distance; it can never be larger.

A commuter train travels west at 90 km/h.

East is positive.

The train brakes with a constant acceleration of 0.50 m/s² directed east.

Calculate the train's velocity after 30 s of braking, and the time it takes to stop.

Step 1: Convert the speed: $90\ \text{km/h} = 90 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 25$ m/s.

The train moves west, so $v_{x0} = -25$ m/s.

Step 2: The acceleration points east: $a_x = +0.50$ m/s².

Step 3: $v_x = v_{x0} + a_xt = -25\ \text{m/s} + (0.50\ \text{m/s}^2)(30\ \text{s}) = -10$ m/s.

Step 4: The train stops when $v_x = 0$: $0 = -25 + 0.50t$, so $t = 50$ s.

Answer: after 30 s the velocity is $-10$ m/s, which is 10 m/s west; the train stops 50 s after it starts braking.

The velocity is still negative (still moving west) but its magnitude fell from 25 m/s to 10 m/s, so the train is slowing down even though its acceleration is positive.

A cart on a level track is moving to the left with speed $v_0$ at time $t = 0$.

A fan on the cart gives it a constant acceleration of magnitude $a$ directed to the right.

Derive an expression for the time $t_s$ at which the cart is momentarily at rest.

Express your final answer in terms of $v_0$, $a$, and physical constants, as appropriate.

Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

$v_x = v_{x0} + a_xt$, right positive (principle line)

$v_{x0} = -v_0$ (the cart starts moving left) and $a_x = +a$ (the acceleration points right)

At $t = t_s$ the cart is at rest, so $v_x = 0$: $0 = -v_0 + at_s$

$t_s = \frac{v_0}{a}$

• Point A1: For a multistep derivation that includes $v_x = v_{x0} + a_xt$.

• Point A2: For opposite signs for $v_{x0}$ and $a_x$, with $v_x = 0$ at the stop.

• Point A3: For the correct expression. Scoring note: a correct, isolated final expression earns Points A2 and A3.

Sense check: the unit is $\frac{\text{m/s}}{\text{m/s}^2} = \text{s}$.

A faster start (bigger $v_0$) takes longer to stop; a stronger fan (bigger $a$) stops it sooner.

At $t = 2t_s$ the same equation gives $v_x = -v_0 + a\frac{2v_0}{a} = +v_0$: the cart is back to its starting speed, now moving right.

Writing $v_{x0} = +v_0$ by habit would give a negative time, a sure sign of a sign error.

A toy car on a straight track has velocity component $v_x = -4.0$ m/s at $t = 0$ and a constant acceleration component $a_x = +2.0$ m/s².

Indicate whether the car's speed at $t = 3.0$ s is greater than, less than, or equal to its speed at $t = 0$.

Justify your answer.

Full-credit answer: "Less than.

At $t = 0$ the velocity is negative and the acceleration is positive, so they point in opposite directions and the car slows down.

It stops when $-4.0 + 2.0t = 0$, at $t = 2.0$ s, and then speeds up in the positive direction.

At $t = 3.0$ s its velocity is $-4.0 + (2.0)(3.0) = +2.0$ m/s, so its speed is 2.0 m/s, which is less than the 4.0 m/s it had at $t = 0$."

• Point A1: For "less than" with an attempt at a relevant justification.

• Point A2: For recognizing that opposite signs of velocity and acceleration mean the car first slows down and reverses.

• Point A3: For comparing the magnitudes of the two velocities (2.0 m/s against 4.0 m/s), not their signed values.

A weaker answer: "Greater than, because the acceleration is positive, so the car speeds up."

This earns nothing.

It treats the sign of the acceleration as if it said "speeding up".

Another weaker answer: "Greater than, because $+2.0$ is bigger than $-4.0$."

This compares signed velocities, not speeds; speed is the magnitude.

A student reads a sensor display that shows a car's acceleration as $-3.0$ m/s² and says, "The car is slowing down."

Describe what further information is needed to decide whether the student is correct, and how it decides the answer.

Full-credit answer: "The sign of the acceleration only shows its direction relative to the chosen positive direction.

To know whether the car is slowing down, we also need the direction of the car's velocity in the same coordinate system.

If the velocity is positive, it is opposite to the acceleration and the car is slowing down.

If the velocity is negative, it is in the same direction as the acceleration and the car is speeding up."

• Point A1: For saying that the direction (sign) of the velocity is needed.

• Point A2: For the rule: same direction means speeding up, opposite directions mean slowing down.

A weaker answer: "We need to know the speed."

The size of the speed alone cannot decide it; it is the direction of the velocity, compared with the direction of the acceleration, that matters.

This earns no point.