Derivatives/Integrals in Motion - College Physics

1. An object starts from rest at position $x_0 = -5$ m. Its acceleration is constant at $a = 2$ m/s². At what time will the object first pass through the origin?
   Answer: The object reaches the origin at $t = \sqrt{5}$ s, when its displacement equals the initial position magnitude. Using the kinematic equation $x(t) = x_0 + v_0 t + \frac{1}{2}at^2$ with $x_0 = -5$ m, $v_0 = 0$ (starts from rest), and $a = 2$ m/s²: $x(t) = -5 + 0 + t^2 = -5 + t^2$. The object reaches the origin when $x(t) = 0$: $-5 + t^2 = 0$, so $t^2 = 5$ and $t = \sqrt{5}$ s ≈ 2.24 s. Choice B gives an incorrect numerical value. Choice C confuses the setup and gives $t = \sqrt{10}$. Choice D uses incorrect reasoning about distance traveled.
2. The acceleration of an object is $a(t) = 4 - 2t$ m/s². If at $t = 1$ s the velocity is 6 m/s, what is the velocity at $t = 3$ s?
   Answer: $v = 6$ m/s When you encounter acceleration as a function of time, you need to integrate to find velocity. Since acceleration is the derivative of velocity with respect to time, $a(t) = \frac{dv}{dt}$, you can find velocity by integrating the acceleration function. Starting with $a(t) = 4 - 2t$, integrate to get: $v(t) = \int (4 - 2t) dt = 4t - t^2 + C$ To find the constant of integration $C$, use the given condition: at $t = 1$ s, $v = 6$ m/s. Substituting: $6 = 4(1) - (1)^2 + C = 4 - 1 + C = 3 + C$ Therefore, $C = 3$, giving us $v(t) = 4t - t^2 + 3$ Now find the velocity at $t = 3$ s: $v(3) = 4(3) - (3)^2 + 3 = 12 - 9 + 3 = 6$ m/s The correct answer is A) $v = 6$ m/s. Looking at the wrong answers: B) $v = 10$ m/s likely comes from incorrectly adding the acceleration at $t = 3$ (which is $4 - 2(3) = -2$) to some intermediate calculation. C) $v = 2$ m/s might result from forgetting the constant of integration or making sign errors. D) $v = 8$ m/s could come from miscalculating the integration constant or arithmetic errors in the final substitution. Remember: when given acceleration as a function of time, always integrate to find velocity, then use initial conditions to determine the integration constant. Double-check your arithmetic, especially when dealing with quadratic terms and negative values.
3. Refer to the graph. The graph shows acceleration $a(t)$ versus time $t$ for an object. If the object has initial velocity $v_0 = 2$ m/s at $t = 0$, what is its velocity at $t = 4$ s?
   Answer: $v = 4$ m/s Velocity is found by integrating acceleration (area under the a-t curve) plus initial velocity. From the graph: $t = 0$ to $t = 2$ s: rectangular area = $(2)(1) = 2$ m/s. $t = 2$ to $t = 4$ s: rectangular area = $(2)(0) = 0$ m/s. Total change in velocity = $2 + 0 = 2$ m/s. Final velocity = $v_0 + \Delta v = 2 + 2 = 4$ m/s. Choice B adds an extra 2 m/s. Choice C assumes no change in velocity. Choice D doubles the correct answer. Choice E subtracts instead of adding the change.
4. The acceleration of an object is constant at $a = -2$ m/s². At $t = 0$, the object has velocity $v_0 = 8$ m/s and position $x_0 = 5$ m. What is the object's position when its velocity becomes zero?
   Answer: $x = 21$ m This is a classic kinematics problem involving constant acceleration. When you see questions with constant acceleration, position, velocity, and time, immediately think of the kinematic equations that relate these variables. To find where the object is when its velocity becomes zero, you need to determine both when this happens and how far it travels. Start with the velocity equation: $v = v_0 + at$. Setting $v = 0$: $0 = 8 + (-2)t$, so $t = 4$ seconds. Now use the position equation: $x = x_0 + v_0t + \frac{1}{2}at^2$. Substituting the known values: $x = 5 + 8(4) + \frac{1}{2}(-2)(4)^2 = 5 + 32 + \frac{1}{2}(-2)(16) = 5 + 32 - 16 = 21$ m. This confirms answer A. Let's examine why the other answers are wrong. Answer B (16 m) likely comes from forgetting the initial position and calculating only $8(4) - 16 = 16$. Answer C (13 m) might result from sign errors or incorrectly handling the acceleration term. Answer D (9 m) could come from various calculation mistakes, possibly confusing the time or acceleration values. The key strategy here is remembering that you have multiple kinematic equations at your disposal. Sometimes it's more efficient to find an intermediate value (like time) first, then use it to find what you're actually looking for. Always double-check your signs—negative acceleration means the object is slowing down in the positive direction.
5. A particle's velocity components are $v_x(t) = 3t$ and $v_y(t) = 4t$ m/s. If the particle starts at the origin, what is its distance from the origin at $t = 2$ s?
   Answer: 10 m This problem tests your ability to connect velocity and position in two-dimensional kinematics. When you're given velocity components as functions of time, you need to integrate to find position, then use the distance formula to find how far the particle has traveled from its starting point. Since velocity is the derivative of position, you can find the position components by integrating the velocity components. For $v_x(t) = 3t$, integrating gives $x(t) = \frac{3t^2}{2} + C_x$. Since the particle starts at the origin, $C_x = 0$, so $x(t) = \frac{3t^2}{2}$. Similarly, for $v_y(t) = 4t$, you get $y(t) = \frac{4t^2}{2} = 2t^2$. At $t = 2$ s: $x(2) = \frac{3(4)}{2} = 6$ m and $y(2) = 2(4) = 8$ m. The distance from the origin is $d = \sqrt{x^2 + y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ m. Answer A (10 m) is correct. Answer B (14 m) likely comes from adding the coordinates instead of using the Pythagorean theorem: $6 + 8 = 14$. Answer C (7 m) might result from incorrectly averaging the velocity components or making integration errors. Answer D (12 m) could come from using the velocity components directly at $t = 2$ without integrating: $\sqrt{6^2 + 8^2}$ but with calculation mistakes. Remember: velocity problems requiring distance always need integration first, then the Pythagorean theorem for the final distance calculation.
6. A particle moves along a line such that its velocity is $v(t) = \sin(\pi t)$ m/s. If the particle is at $x = 2$ m when $t = 0$, what is its position at $t = 1$ s?
   Answer: $x = 2 + \frac{2}{\pi}$ m When you encounter a problem involving velocity and position, remember that velocity is the derivative of position with respect to time. To find position from velocity, you need to integrate and apply initial conditions. Given $v(t) = \sin(\pi t)$, you find position by integrating: $x(t) = \int \sin(\pi t) \, dt = -\frac{1}{\pi}\cos(\pi t) + C$. To find the constant $C$, use the initial condition $x(0) = 2$: $2 = -\frac{1}{\pi}\cos(0) + C = -\frac{1}{\pi} + C$, so $C = 2 + \frac{1}{\pi}$. Therefore, $x(t) = -\frac{1}{\pi}\cos(\pi t) + 2 + \frac{1}{\pi}$. At $t = 1$: $x(1) = -\frac{1}{\pi}\cos(\pi) + 2 + \frac{1}{\pi} = -\frac{1}{\pi}(-1) + 2 + \frac{1}{\pi} = \frac{1}{\pi} + 2 + \frac{1}{\pi} = 2 + \frac{2}{\pi}$. Answer A is correct. Answer B ($x = 2$) assumes the particle doesn't move at all, ignoring the velocity function entirely. Answer C ($x = 2 - \frac{2}{\pi}$) likely results from a sign error when integrating or applying $\cos(\pi) = -1$. Answer D ($x = \frac{2}{\pi}$) forgets to include the initial position offset and only considers the displacement. Remember: when integrating to find position from velocity, always apply your initial conditions carefully to determine the constant of integration. The particle's final position equals its initial position plus the net displacement over the time interval.
7. An object's position is described by $x(t) = 4t^2 - 12t + 8$ meters. What is the object's average velocity between $t = 1$ s and $t = 4$ s?
   Answer: $\bar{v} = 8$ m/s When you encounter a position function and need to find average velocity, remember that average velocity is the total displacement divided by the time interval, not the average of instantaneous velocities. The formula for average velocity is $\bar{v} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$, where you need the position at both endpoints of the time interval. First, find the position at each time: - At $t = 1$ s: $x(1) = 4(1)^2 - 12(1) + 8 = 4 - 12 + 8 = 0$ m - At $t = 4$ s: $x(4) = 4(4)^2 - 12(4) + 8 = 64 - 48 + 8 = 24$ m Now calculate: $\bar{v} = \frac{24 - 0}{4 - 1} = \frac{24}{3} = 8$ m/s This confirms answer A is correct. Answer B (4 m/s) likely comes from incorrectly calculating the displacement or making an arithmetic error. Answer C (12 m/s) might result from confusing average velocity with the coefficient from the velocity function $v(t) = 8t - 12$, or from using $|12|$ from the original equation. Answer D (6 m/s) could come from averaging the instantaneous velocities at the endpoints: $v(1) = -4$ m/s and $v(4) = 20$ m/s gives $\frac{-4 + 20}{2} = 8$ m/s, but this approach is conceptually wrong for average velocity. Remember: average velocity depends only on initial and final positions, not on the path taken or instantaneous velocities along the way.
8. The position of a particle is given by $x(t) = At^3 + Bt^2 + Ct + D$, where $A$, $B$, $C$, and $D$ are constants. If the particle has zero acceleration at $t = 1$ s, what condition must be satisfied?
   Answer: $6A + 2B = 0$ When you encounter position-time functions in physics, remember that velocity and acceleration are found by taking successive derivatives. This question tests your understanding of the relationship between these kinematic quantities. Given the position function $x(t) = At^3 + Bt^2 + Ct + D$, you need to find the acceleration by differentiating twice. The velocity is the first derivative: $v(t) = \frac{dx}{dt} = 3At^2 + 2Bt + C$. The acceleration is the second derivative: $a(t) = \frac{dv}{dt} = 6At + 2B$. Since the particle has zero acceleration at $t = 1$ s, you substitute this value: $a(1) = 6A(1) + 2B = 0$, which gives you $6A + 2B = 0$. This confirms that answer A is correct. Now let's examine why the other options are wrong. Answer B ($3A + 2B + C = 0$) represents the velocity equation at $t = 1$, not the acceleration condition. Answer C ($A + B + C + D = 0$) would be the position at $t = 1$, which is unrelated to the acceleration requirement. Answer D ($3A + B = 0$) appears to be a partial or incorrectly derived acceleration condition, missing the proper coefficient for the $B$ term. Remember this pattern: always take the second derivative to find acceleration from a position function. Many students mistakenly use the first derivative (velocity) or substitute incorrectly. Practice identifying which kinematic quantity the question asks for before you start differentiating.

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