Representing Motion

Choose downward as positive. The ball has zero initial vertical velocity. Equation used: $$y={y}_{0}+{v}_{y0}t+\frac{1}{2}{a}_{y}{t}^{2}$$ Rearrange for $t$: $$t=\sqrt{\frac{2(y-{y}_{0})}{{a}_{y}}}$$ Given: $$y-{y}_{0}=45\text{ m}$$ $${v}_{y0}=0\text{ m/s}$$ $${a}_{y}=10{\text{ m/s}}^{2}$$ Working: $$t=\sqrt{\frac{2(45)}{10}}$$ $$t=\sqrt{9.0}$$ $$t=3.0\text{ s}$$

Greater than. Because $$x={x}_{0}+\frac{1}{2}{a}_{x}{t}^{2}$$ for motion from rest, displacement grows as ${t}^{2}$. In the first 3.0 s the cyclist covers $$\frac{1}{2}(2.0)(3.0{)}^{2}=9.0\text{ m}$$ In 6.0 s total, $$\frac{1}{2}(2.0)(6.0{)}^{2}=36\text{ m}$$ The second interval therefore contributes $$36-9.0=27\text{ m}$$ which is greater than 9.0 m. The squared dependence on time means each successive equal time interval produces a larger displacement.

Choose downward as positive. Velocity: $${v}_{y}={v}_{y0}+{a}_{y}t$$ $${v}_{y}=0+(10)(4.0)$$ $${v}_{y}=40\text{ m/s}$$ Distance: $$y-{y}_{0}={v}_{y0}t+\frac{1}{2}{a}_{y}{t}^{2}$$ $$y-{y}_{0}=0+\frac{1}{2}(10)(4.0{)}^{2}$$ $$y-{y}_{0}=80\text{ m}$$ The stone reaches 40 m/s and falls 80 m.

The same time. Free-fall acceleration $g$ does not depend on mass. Both stones experience the same ${a}_{y}$, start from rest, and fall the same height. The kinematic equations contain no mass term, so both reach the ground simultaneously.

Starting point: Definition of instantaneous velocity. $${v}_{x}=\frac{dx}{dt}$$ Step 1 — Differentiate $x(t)$ using the power rule $${v}_{x}=\frac{d}{dt}(3{t}^{3}-5t+2)=9{t}^{2}-5\text{ m/s}$$ Step 2 — Differentiate ${v}_{x}(t)$ to find acceleration $${a}_{x}=\frac{d{v}_{x}}{dt}=\frac{d}{dt}(9{t}^{2}-5)=18t{\text{ m/s}}^{2}$$ The velocity is ${v}_{x}(t)=9{t}^{2}-5$ m/s and the acceleration is ${a}_{x}(t)=18t$ m/s².

Starting point: Change in velocity equals the integral of acceleration. $$\Delta {v}_{x}=\int_{0}^{t}{a}_{x}({t}^{′}) d{t}^{′}$$ Step 1 — Substitute the acceleration function $$\Delta {v}_{x}=\int_{0}^{t}6{t}^{′} d{t}^{′}$$ Step 2 — Evaluate using the power rule $$\Delta {v}_{x}={\left[3{t}^{′}{}^{2}\right]}_{0}^{t}=3{t}^{2}$$ Because ${v}_{x0}=0$, the velocity is ${v}_{x}(t)=3{t}^{2}$. At $t=2.0$ s: $${v}_{x}=3(2.0{)}^{2}=12\text{ m/s}$$

The claim is incorrect. Acceleration is the derivative of velocity: ${a}_{x}=\frac{d{v}_{x}}{dt}$. A zero velocity means the slope of the $x$-vs-$t$ graph is zero at that instant. However, acceleration depends on how velocity changes, not on the value of velocity itself. At the peak, velocity changes from positive (upward) to negative (downward), so $\frac{d{v}_{y}}{dt}$ is nonzero. The acceleration remains $g\approx 10{\text{ m/s}}^{2}$ downward throughout free fall.

Consistent. The particle changes direction when ${v}_{x}=0$, which occurs at $t=4.0$ s. From $t=0$ to $t=4.0$ s the velocity is positive, so the particle moves in the positive direction. From $t=4.0$ s to $t=6.0$ s the velocity is negative, so the particle moves in the negative direction. The net displacement of 12 m is less than the distance traveled in the positive direction alone ($\int_{0}^{4}(8-2t) dt=16$ m), confirming that some backward displacement ($-4$ m) partially canceled the forward displacement. This is consistent with a direction change.