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An object moves from ${x}_{0}=3$ m to $x=8$ m, then back to $x=5$ m. Determine the displacement and the total distance traveled.
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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.
Indicate whether the magnitude of displacement is greater than, less than, or equal to the total distance traveled for an object that reverses direction? Justify your response.
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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.
A ball rolls from $x=2$ m to $x=14$ m in 4.0 s. Calculate the average velocity.
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$${\vec{v}}_{\text{avg}}=\frac{\Delta x}{\Delta t}$$ $$\Delta x=14-2=12\text{ m}$$ $${\vec{v}}_{\text{avg}}=\frac{12}{4.0}=3.00\text{ m/s}$$ The average velocity is 3.00 m/s in the positive $x$-direction.
A car traveling at 25 m/s north brakes to 5.0 m/s north in 8.0 s. Determine the average acceleration and describe its direction relative to the velocity?
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$${\vec{a}}_{\text{avg}}=\frac{\Delta v}{\Delta t}$$ $${\vec{a}}_{\text{avg}}=\frac{5.0-25}{8.0}=\frac{-20}{8.0}$$ $${\vec{a}}_{\text{avg}}=-2.50{\text{ m/s}}^{2}$$ The negative sign means the acceleration is directed south — opposite to the northward velocity. The car is decelerating.
Indicate whether the average velocity over a round trip (returning to the starting point) is greater than, less than, or equal to zero? Justify your response.
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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.
The position of a particle is given by $x(t)=4{t}^{2}-3t+7$ (SI units). Derive expressions for the instantaneous velocity and acceleration as functions of time.
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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.
A particle has velocity ${v}_{x}(t)=12{t}^{2}+5$ m/s and starts at $x(0)=0$. Derive an expression for $x(t)$ by setting up and evaluating the appropriate integral.
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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: $$x(t)=4{t}^{3}+5t\text{ m}$$
Indicate whether the instantaneous velocity at $t=2$ s is greater than, less than, or equal to the average velocity over the interval $0\le t\le 4$ s for the function $x(t)={t}^{3}$? Justify your response.
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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.
Describe how the acceleration-time graph for $x(t)=5{t}^{3}-2t+1$ would differ if the position function were changed to $x(t)=5{t}^{2}-2t+1$? Briefly justify your answer.
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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.