Kinematics All subtopics

Finding the distance traveled $$d=200+150=350\text{ m}$$ Finding the displacement (taking north as positive) $$s=(+200)+(-150)=+50\text{ m}$$ The distance traveled is 350 m and the displacement is 50 m north.

Speed cannot be negative because speed is a scalar quantity defined as the magnitude of velocity. Magnitude is always zero or positive, so speed has no directional sign.

Taking rightward as positive, the components are +8 N and −12 N. Finding the vector sum $${F}_{\text{net}}=(+8)+(-12)=-4\text{ N}$$ The negative sign means the resultant force is 4 N to the left.

Displacement can be zero when distance is nonzero. Displacement depends only on the initial and final positions, so if an object returns to its starting point, the change in position is zero. Distance, however, accumulates every meter of path traveled, so it remains positive for any object that has moved at all.

Equation used: $$\vec{d}={x}_{f}-{x}_{i}$$ Given: $${x}_{f}=3\text{ m}$$ $${x}_{i}=8\text{ m}$$ Substitution: $$\vec{d}=3-8$$ $$\vec{d}=-5\text{ m}$$ The displacement is −5 m. The negative sign indicates the object moved 5 m in the negative direction.

The displacement is zero because the runner’s final position is identical to her initial position, so the change in position is zero. The distance travelled is 400 m because distance measures the total path length. Displacement depends only on the start and end positions, while distance accumulates every metre of the path, so they differ whenever the path is not a straight line in one direction.

Converting distance to SI: $$90\text{ km}=90\times 1000=90,000\text{ m}$$ $$30\text{ km}=30\times 1000=30,000\text{ m}$$ Converting time to SI: $$1.5\text{ h}=1.5\times 3600=5400\text{ s}$$ $$0.5\text{ h}=0.5\times 3600=1800\text{ s}$$ Finding total displacement: $$\vec{d}=(+90,000)+(-30,000)=+60,000\text{ m}$$ Finding total time: $$\Delta t=5400+1800=7200\text{ s}$$ Equation used: $${\vec{v}}_{avg}=\frac{\vec{d}}{\Delta t}$$ Substitution: $${\vec{v}}_{avg}=\frac{60,000}{7200}$$ $${\vec{v}}_{avg}=+8.33\text{ m/s}$$ The average velocity is +8.33 m/s (north).

The magnitude of average velocity is less than the average speed. For a round trip the displacement is zero, so average velocity is zero. The total distance is nonzero because the object moved along a path, so average speed is positive. Zero is less than any positive value, so the magnitude of average velocity is less than the average speed.

Equation used: $${\vec{a}}_{avg}=\frac{{\vec{v}}_{f}-{\vec{v}}_{i}}{\Delta t}$$ Given: $${\vec{v}}_{f}=+10.0\text{ m/s}$$ $${\vec{v}}_{i}=+4.0\text{ m/s}$$ $$\Delta t=2.0\text{ s}$$ Substitution: $${\vec{a}}_{avg}=\frac{10.0-4.0}{2.0}$$ $${\vec{a}}_{avg}=\frac{6.0}{2.0}$$ $${\vec{a}}_{avg}=+3.0{\text{ m/s}}^{2}$$ The positive sign confirms the acceleration is in the same direction as the velocity, so the ball speeds up.

The satellite’s acceleration is nonzero. Although its speed is constant, the direction of its velocity continuously changes as it follows the circular path. Because acceleration occurs whenever the magnitude and/or direction of velocity changes, a change in direction alone is sufficient for nonzero acceleration.

The object is speeding up. The velocity and the acceleration are both in the negative direction (same sign). When velocity and acceleration point in the same direction, the speed increases. The magnitude of velocity grows from 6.0 m/s to a larger value over time.

The student draws a tangent line to the curve at $t=3$ s. The tangent line touches the curve at exactly one point. The student then calculates the slope of that tangent line by choosing two widely spaced points on the line and dividing the rise by the run. That slope equals the instantaneous velocity at $t=3$ s.

The answer cannot be determined without knowing whether the curve is concave up or concave down. If the curve is concave up (accelerating), the instantaneous velocity at $t=5$ s is less than the average velocity over the 0–10 s interval because the object moves slower in the first half and faster in the second half. If the curve is concave down (decelerating), the opposite is true. The shape of the graph must be specified before a comparison can be made.

For the ball slowing down, the dots are far apart on the left and become closer together toward the right, because the speed decreases over time. The velocity arrows point to the right but get shorter at each successive dot. The acceleration arrows point to the left — opposite to the velocity — because the ball is decelerating. For the ball speeding up, the pattern reverses: dots spread apart toward the right, velocity arrows grow longer, and acceleration arrows point to the right, in the same direction as the velocity.

An equation gives the mathematical relationship among position, velocity, acceleration, and time, but it does not show the physical context or the direction of travel in a real scenario. A narrative or diagram can clarify where the object starts, which way it moves, and what forces act on it. Combining representations gives a more complete and less ambiguous picture of the motion.

Taking the initial direction of travel as positive: Equation used: $${v}_{x}^{2}={v}_{x0}^{2}+2{a}_{x}(x-{x}_{0})$$ Rearranging for displacement: $$x-{x}_{0}=\frac{{v}_{x}^{2}-{v}_{x0}^{2}}{2{a}_{x}}$$ Given: $${v}_{x}=0\text{ m/s}$$ $${v}_{x0}=25\text{ m/s}$$ $${a}_{x}=-5.0{\text{ m/s}}^{2}$$ Working — substitution: $$x-{x}_{0}=\frac{(0{)}^{2}-(25{)}^{2}}{2(-5.0)}$$ $$x-{x}_{0}=\frac{-625}{-10}$$ $$x-{x}_{0}=62.5\text{ m}$$ The car travels 62.5 m before coming to rest. The positive result confirms the car moves in the original direction of travel while stopping.

Doubling the braking time does not simply double the displacement. The position–time equation shows that displacement equals ${v}_{x0}t+\frac{1}{2}{a}_{x}{t}^{2}$. The $\frac{1}{2}{a}_{x}{t}^{2}$ term is proportional to ${t}^{2}$, so it quadruples when $t$ doubles. The ${v}_{x0}t$ term does double, but the combined result is not a simple doubling. Therefore the displacement more than doubles if ${a}_{x}$ acts in the same direction as ${v}_{x0}$, or increases by less than double if ${a}_{x}$ opposes ${v}_{x0}$.

Derivation: Time to travel displacement $d$ from rest Starting point: The position–time kinematic equation. $$x={x}_{0}+{v}_{x0}t+\frac{1}{2}{a}_{x}{t}^{2}$$ Step 1 — Define displacement and apply the initial condition ${v}_{x0}=0$. $$d=(0)t+\frac{1}{2}a {t}^{2}$$ $$d=\frac{1}{2}a {t}^{2}$$ Step 2 — Isolate ${t}^{2}$. $${t}^{2}=\frac{2d}{a}$$ Step 3 — Take the positive square root (time must be positive). $$t=\sqrt{\frac{2d}{a}}$$ Result: $$t=\sqrt{\frac{2d}{a}}$$ The time is proportional to the square root of displacement. Quadrupling the displacement only doubles the required time.

Taking upward as positive: Equation used: $${v}_{y}^{2}={v}_{y0}^{2}+2{a}_{y}(y-{y}_{0})$$ At maximum height ${v}_{y}=0$. Rearranging for displacement: $$y-{y}_{0}=\frac{{v}_{y}^{2}-{v}_{y0}^{2}}{2{a}_{y}}$$ Given: $${v}_{y}=0\text{ m/s}$$ $${v}_{y0}=20\text{ m/s}$$ $${a}_{y}=-10{\text{ m/s}}^{2}$$ Working — substitution: $$y-{y}_{0}=\frac{(0{)}^{2}-(20{)}^{2}}{2(-10)}$$ $$y-{y}_{0}=\frac{-400}{-20}$$ $$y-{y}_{0}=20\text{ m}$$ The maximum height reached by the ball is 20 m.

The ball thrown at $20\text{ m/s}$ spends more total time in the air. The time to reach the peak is $t=\frac{{v}_{y}-{v}_{y0}}{{a}_{y}}$. A larger initial velocity means a longer time to decelerate to zero. Since the descent mirrors the ascent, the total flight time is proportional to ${v}_{y0}$. Therefore the $20\text{ m/s}$ ball is airborne for twice as long as the $10\text{ m/s}$ ball.