Representing Motion

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.

The maximum height would quadruple. From ${v}_{y}^{2}={v}_{y0}^{2}+2{a}_{y}(y-{y}_{0})$ with ${v}_{y}=0$, the height is $y-{y}_{0}=\frac{{v}_{y0}^{2}}{2g}$. Height is proportional to ${v}_{y0}^{2}$. Doubling ${v}_{y0}$ means ${v}_{y0}^{2}$ increases by a factor of four, so the height increases by a factor of four — from 20 m to 80 m.

The displacement equals the area under the velocity–time graph. The shape is a trapezoid. Equation used — area of a trapezoid: $$\text{displacement}=\frac{1}{2}({v}_{1}+{v}_{2})\times t$$ Given: $${v}_{1}=4\text{ m/s}$$ $${v}_{2}=12\text{ m/s}$$ $$t=4\text{ s}$$ Working — substitution: $$\text{displacement}=\frac{1}{2}(4+12)(4)$$ $$\text{displacement}=\frac{1}{2}(16)(4)$$ $$\text{displacement}=32\text{ m}$$ The displacement of the object during this interval is 32 m.

Draw a tangent line to the velocity–time curve at $t=3\text{ s}$. Choose two points on the tangent line that are far apart. Calculate the slope by dividing the change in velocity by the change in time between those two points. That slope equals the instantaneous acceleration at $t=3\text{ s}$.

The statement is not consistent with kinematics. An object can have a positive velocity while having a negative acceleration — it moves in the positive direction but is slowing down. The sign of acceleration indicates the direction of the net change in velocity, not the direction of travel. A car moving right (positive velocity) while braking has a negative acceleration.

The position–time graph is a downward-opening parabola (concave down) that curves away from the origin in the negative-position direction. The object starts at rest, so the tangent to the curve at $t=0$ is horizontal (zero slope). As time increases, the slope becomes increasingly negative, reflecting the growing speed in the negative direction. The acceleration is constant and negative, so the velocity–time graph is a straight line with a negative slope passing through the origin.