Instantaneous velocity is the first derivative of position with respect to time, $$v(t) = \frac{dx}{dt}$$. Instantaneous acceleration is the first derivative of velocity with respect to time, $$a(t) = \frac{dv}{dt}$$. Substituting the expression for velocity gives $$a(t) = \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{d^2x}{dt^2}$$.

Displacement is the integral of the velocity function over the time interval. $$\Delta x = \int_{1}^{3} v(t) dt = \int_{1}^{3} (3t^2 + 2t) dt = [t^3 + t^2]_{1}^{3} = (3^3 + 3^2) - (1^3 + 1^2) = (27 + 9) - (1 + 1) = 36 - 2 = 34$$ m.

We must integrate twice. First, find velocity: $$v(t) = \int a(t) dt = \int kt dt = \frac{1}{2}kt^2 + C_1$$. Since $$v(0)=0$$, $$C_1=0$$. Next, find position: $$x(t) = \int v(t) dt = \int \frac{1}{2}kt^2 dt = \frac{1}{6}kt^3 + C_2$$. Since $$x(0)=0$$, $$C_2=0$$. Thus, $$x(t) = \frac{1}{6}kt^3$$.

Using the standard kinematic equation for position under constant acceleration: $$y(t) = y_0 + v_{0y}t + \frac{1}{2}a_y t^2$$. Here, the initial position $$y_0 = 0$$, the initial velocity $$v_{0y} = v_0$$, and the acceleration $$a_y = -g$$ because gravity acts downward. Substituting these values gives $$y(t) = v_0 t - \frac{1}{2}gt^2$$.

This question tests AP Physics C: Mechanics skills in representing motion through vertical projectile analysis. Vertical motion with constant acceleration follows kinematic equations where maximum height occurs when velocity reaches zero. In this scenario, a rocket launches upward at 18 m/s against gravity (-9.8 m/s²), reaching maximum height when v = 0, using v² = v0² - 2gh gives h = v0²/(2g) = (18)²/(2×9.8) ≈ 16.5 m. Choice B is correct because it properly calculates the maximum height using energy conservation or kinematic equations. Choice D is incorrect because it confuses height (measured in meters) with velocity (measured in m/s), a common dimensional error when students rush through problems. To help students: Emphasize checking units in final answers. Practice recognizing when velocity equals zero at turning points in motion.

This question tests AP Physics C: Mechanics skills in representing motion through relative motion analysis. When objects move with constant velocities, displacement equals velocity times time for each object independently. In this scenario, Cart 1 moves at +2.0 m/s for 5 seconds, giving displacement Δx1 = 2.0 × 5 = +10 m, while Cart 2 moves at -1.0 m/s for 5 seconds, giving displacement Δx2 = -1.0 × 5 = -5 m. Choice A is correct because it properly calculates displacement as velocity times time for each cart with correct signs. Choice D is incorrect because it confuses displacement (measured in meters) with velocity (measured in m/s), showing the velocity values instead of calculating displacement. To help students: Emphasize the difference between instantaneous quantities (velocity) and accumulated quantities (displacement). Practice problems with multiple objects to build comfort with relative motion.

This question tests AP Physics C: Mechanics skills in representing motion through kinematic equation application. One-dimensional motion with constant acceleration follows position and velocity equations derived from calculus or kinematic relationships. In this scenario, with y0 = 1.5 m, v0 = 14 m/s, and a = -9.8 m/s², the position equation is y(t) = y0 + v0t + ½at² = 1.5 + 14t - 4.9t², and velocity is v(t) = v0 + at = 14 - 9.8t. Choice A is correct because it properly applies both kinematic equations with correct signs for upward initial velocity and downward acceleration. Choice B is incorrect because it uses positive acceleration (+4.9t²), failing to recognize that gravity acts downward (negative in the upward-positive convention). To help students: Consistently define coordinate systems before solving. Practice deriving velocity from position equations to verify consistency.

This question tests AP Physics C: Mechanics skills in representing motion through parametric equations for projectile motion. Projectile motion requires separate kinematic equations for horizontal and vertical components, with constant horizontal velocity and constant vertical acceleration. In this scenario, a ball is thrown at an angle, requiring component analysis with x(t) for horizontal motion and y(t) for vertical motion. Choice A is correct because horizontal motion has constant velocity: x(t)=(v₀cosθ)t with no acceleration term, while vertical motion includes initial velocity and gravitational acceleration: y(t)=(v₀sinθ)t-½gt². Choice B is incorrect because it incorrectly includes ½ in the horizontal equation, suggesting acceleration where none exists, and omits the ½ factor in the vertical equation. To help students: Emphasize that horizontal motion has zero acceleration in projectile problems. Create tables showing initial conditions and accelerations for each component, and practice deriving position equations from first principles.

This question tests AP Physics C: Mechanics skills in representing motion through projectile motion analysis. Projectile motion involves analyzing the independent horizontal and vertical components of motion, where only gravity acts in the vertical direction. In this scenario, a soccer ball is kicked at 20 m/s at 35° above horizontal, creating initial velocity components vx = 20cos(35°) ≈ 16.4 m/s and vy = 20sin(35°) ≈ 11.5 m/s. Choice B is correct because the maximum height occurs when vy = 0, using vy² = v0y² - 2gh gives h = v0y²/(2g) = (11.5)²/(2×9.8) ≈ 6.6 m. Choice C is incorrect because it doubles the correct answer, a common error when students confuse the maximum height formula with total vertical displacement. To help students: Use component analysis diagrams to visualize velocity vectors. Practice identifying when vertical velocity equals zero at the peak of trajectory.

This question tests AP Physics C: Mechanics skills in representing motion through kinematic analysis of projectile motion. Projectile motion involves analyzing the independent horizontal and vertical components of motion, where the vertical component experiences constant acceleration due to gravity while the horizontal component maintains constant velocity. In this scenario, a soccer ball is kicked at 20 m/s at a 35° angle, requiring us to find the maximum height using kinematic equations. Choice B is correct because at maximum height, the vertical velocity becomes zero, so using v²=v₀²+2aΔy with v_y=0, v₀y=20sin(35°)≈11.5 m/s, and a=-9.8 m/s², we get Δy=v₀y²/(2g)≈6.6 m. Choice C is incorrect because it appears to use half the correct value, possibly from misapplying the equation or using the wrong initial velocity component. To help students: Emphasize decomposing initial velocity into components using trigonometry. Use energy methods as an alternative check, and practice identifying when vertical velocity equals zero at the peak.