Kinematics All subtopics

Find the magnitude: $$|\vec{F}|=\sqrt{(-8{)}^{2}+{6}^{2}}$$ $$|\vec{F}|=\sqrt{64+36}=\sqrt{100}=10\text{ N}$$ Find the reference angle: $${\theta }_{\text{ref}}={tan}^{-1}\left(\frac{6}{8}\right)=36.9^{\circ}$$ The x-component is negative and the y-component is positive. This places the vector in the second quadrant. $$\theta =180^{\circ}-36.9^{\circ}=143^{\circ}\text{ from the +x axis}$$ The force has a magnitude of 10 N directed at 143° from the positive x-axis.

Distance is greater. Distance is a scalar that accumulates every meter traveled, so the total distance is 10 km. Displacement is a vector from start to finish, and because the car returns to its starting point, the displacement is 0 km. A scalar sum of path lengths is always greater than or equal to the magnitude of the net displacement.

Starting point: The resultant vector equals the component-wise sum of the addend vectors. $$\vec{R}=\vec{A}+\vec{B}$$ $$\vec{R}=(3+(-1))\hat{i}+((-2)+5)\hat{j}$$ $$\vec{R}=2\hat{i}+3\hat{j}\text{ m}$$ Find the magnitude: $$|\vec{R}|=\sqrt{{2}^{2}+{3}^{2}}=\sqrt{4+9}=\sqrt{13}$$ $$|\vec{R}|\approx 3.61\text{ m}$$

The claim is not consistent. Speed and velocity share the same SI unit, but they are fundamentally different. Speed is a scalar - it has magnitude only and is always non-negative. Velocity is a vector - it has both magnitude and direction. A car moving at 20 m/s west and another at 20 m/s east have the same speed but different velocities. The unit alone does not determine whether a quantity is scalar or vector.

At the highest point, velocity is zero — the ball has momentarily stopped before reversing direction. Acceleration is negative. Gravity acts downward, which is the negative direction in this coordinate system. The acceleration due to gravity is $a=-g=-10$ m/s² regardless of the ball’s position or velocity.

Distance is a scalar, so it accumulates without regard to direction: total distance is 11 m, a positive number. Displacement is a vector from start to finish. The object ends 5 m to the right of where it started, so the displacement is +5 m. The positive sign indicates the net displacement is in the positive (rightward) direction. Distance has no sign; displacement carries a sign that shows direction.

Displacement uses only initial and final positions. $$\Delta x=x-{x}_{0}=5-3$$ $$\Delta x=2\text{ m}$$ The total distance is $5+3=8$ m. Displacement is 2 m in the positive direction, while distance is 8 m.

The magnitude of displacement is less than the total distance traveled. Displacement measures only the net change in position, so the backtracked portion cancels. Distance adds every segment regardless of direction, so it is always at least as large as the magnitude of displacement.

$${\vec{v}}_{\text{avg}}=\frac{\Delta x}{\Delta t}$$ $$\Delta x=14-2=12\text{ m}$$ $${\vec{v}}_{\text{avg}}=\frac{12}{4.0}=3.00\text{ m/s}$$ The average velocity is 3.00 m/s in the positive $x$-direction.

$${\vec{a}}_{\text{avg}}=\frac{\Delta v}{\Delta t}$$ $${\vec{a}}_{\text{avg}}=\frac{5.0-25}{8.0}=\frac{-20}{8.0}$$ $${\vec{a}}_{\text{avg}}=-2.50{\text{ m/s}}^{2}$$ The negative sign means the acceleration is directed south — opposite to the northward velocity. The car is decelerating.

The average velocity equals zero. Displacement for a round trip is zero because the final position equals the initial position. Since ${\vec{v}}_{\text{avg}}=\Delta \vec{x}/\Delta t$ and $\Delta \vec{x}=0$, the average velocity must be zero regardless of the time elapsed.

Starting point: Definition of instantaneous velocity. $${v}_{x}=\frac{dx}{dt}$$ Step 1 — Differentiate $x(t)$ with respect to $t$ $${v}_{x}(t)=\frac{d}{dt}(4{t}^{2}-3t+7)=8t-3\text{ m/s}$$ Step 2 — Differentiate ${v}_{x}(t)$ to find acceleration $${a}_{x}(t)=\frac{d{v}_{x}}{dt}=\frac{d}{dt}(8t-3)=8{\text{ m/s}}^{2}$$ The velocity increases linearly with time. The acceleration is constant at 8 m/s².

Starting point: Position is the integral of velocity. $$x(t)=\int {v}_{x}(t) dt$$ Step 1 — Set up the integral The integration variable is $t$. The integrand $12{t}^{2}+5$ represents the velocity at each instant. $$x(t)=\int (12{t}^{2}+5) dt$$ Step 2 — Evaluate using the power rule $$x(t)=4{t}^{3}+5t+C$$ Step 3 — Apply initial condition $x(0)=0$ $$0=4(0{)}^{3}+5(0)+C⟹C=0$$ Result: $$x(t)=4{t}^{3}+5t\text{ m}$$

The instantaneous velocity at $t=2$ s is less than the average velocity over $[0,4]$ s. $${v}_{x}(2)=3(2{)}^{2}=12\text{ m/s}$$ $${\vec{v}}_{\text{avg}}=\frac{\Delta x}{\Delta t}=\frac{64-0}{4}=16\text{ m/s}$$ Because $x={t}^{3}$ is concave up (accelerating), the function’s slope at the midpoint is less than the secant slope across the full interval.

For $x=5{t}^{3}-2t+1$, the acceleration is ${a}_{x}=30t$, a line through the origin with positive slope. For $x=5{t}^{2}-2t+1$, the acceleration is ${a}_{x}=10$, a horizontal line. The key difference is slope: the $a$-$t$ graph changes from a rising line to a flat line. This occurs because the cubic term produces a linearly increasing acceleration, while the quadratic term produces constant acceleration.

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.