AP Physics C: Mechanics Kinematics — Worked Answer Explanations

Unit 1 · 12 questions explained

1. Question 1 · Easy

   A particle moves along the $x$-axis with velocity $v(t)=3t^2-6t$ m/s. What is the acceleration at $t=2$ s?

   • A
     $6$ m/s²Correct
   • B
     $0$ m/s²

     Why not B: Evaluated $v(2)=12-12=0$ and reported that as the acceleration.
   • C
     $12$ m/s²

     Why not C: Used only the $6t$ term without subtracting $6$.
   • D
     $-6$ m/s²

     Why not D: Treated only the constant $-6$ as the derivative.
   Explanation

   Acceleration is $a(t)=\frac{dv}{dt}=6t-6$. At $t=2$ s: $a(2)=6(2)-6=12-6=6$ m/s².

   Key takeaway

   Differentiate $v(t)$ to get $a(t)$, then evaluate at the given time.

2. Question 2 · Easy

   A particle starts from rest at the origin and accelerates with $a(t)=4t$ m/s². What is the velocity at $t=3$ s?

   • A
     $12$ m/s

     Why not A: Evaluated $a(3)=12$ without integrating to find velocity.
   • B
     $9$ m/s

     Why not B: Computed $\frac{1}{2}(4)(3)=6$ and made an arithmetic error.
   • C
     $18$ m/sCorrect
   • D
     $36$ m/s

     Why not D: Forgot the factor of $\frac{1}{2}$ when integrating $4t$.
   Explanation

   $v(t)=\int adt=\int 4tdt=2t^2+C$. With $v(0)=0$: $C=0$. At $t=3$: $v(3)=2(9)=18$ m/s.

   Key takeaway

   Integrate $a(t)$ and apply the initial condition $v(0)$ to find the constant of integration.

3. Question 3 · Easy

   A ball is launched horizontally from a cliff 80 m high with speed $20$ m/s. Using $g=10$ m/s², how far from the base of the cliff does it land?

   • A
     $40$ m

     Why not A: Used $t=2$ s instead of finding $t$ from the free-fall equation.
   • B
     $80$ mCorrect
   • C
     $160$ m

     Why not C: Used the linear equation $80=10t$ instead of $80=\frac{1}{2}(10)t^2$.
   • D
     $20$ m

     Why not D: Divided height by $g$ without the factor of $\frac{1}{2}$.
   Explanation

   Vertical free-fall: $80=\frac{1}{2}(10)t^2\Rightarrow t^2=16\Rightarrow t=4$ s. Horizontal range: $x=v_{0x}\cdot t=20\times 4=80$ m.

   Key takeaway

   In projectile motion, solve for flight time using vertical kinematics ($y = \frac{1}{2}gt^2$), then multiply by horizontal speed.

4. Question 4 · Easy

   Given $a(t)=6t$ m/s² with $v(0)=0$ and $x(0)=2$ m, find $x(3)$.

   • A
     $27$ m

     Why not A: Forgot to add $x(0)=2$ m.
   • B
     $29$ mCorrect
   • C
     $54$ m

     Why not C: Integrated $a$ only once and used the result as position.
   • D
     $20$ m

     Why not D: Used $x=\frac{1}{2}a{t}^2$ with $a=6$ (a constant) rather than the time-varying value.
   Explanation

   Integrate $a$: $v(t)=\int 6tdt=3t^2+C_1$. With $v(0)=0$, $C_1=0$. Integrate $v$: $x(t)=\int 3t^{2}dt=t^3+C_2$. With $x(0)=2$, $C_2=2$. So $x(3)=27+2=29$ m.

   Key takeaway

   Apply initial conditions after each integration step to determine the constants.

5. Question 5 · Easy

   A particle's position is $x(t)=t^3-3t^2+2$ m. At what time(s) is the particle momentarily at rest for $t\ge 0$?

   • A
     $t=0$ only

     Why not A: Set $x(t)=0$ instead of $v(t)=0$.
   • B
     $t=1$ s only

     Why not B: Found when acceleration is zero ($a=0$ at $t=1$) rather than velocity.
   • C
     $t=0$ and $t=2$ sCorrect
   • D
     $t=2$ s only

     Why not D: Missed the $t=0$ solution.
   Explanation

   Velocity: $v(t)=\frac{dx}{dt}=3t^2-6t=3t(t-2)$. Setting $v=0$: $t=0$ or $t=2$ s.

   Key takeaway

   The particle is at rest when $v(t) = \frac{dx}{dt} = 0$, not when $x(t) = 0$.

6. Question 6 · Medium

   A projectile is launched at angle $\theta$ above the horizontal with initial speed $v_0$. Which expression gives the maximum height?

   • A
     $\frac{v_0^2}{2g}$

     Why not A: Used total initial speed instead of the vertical component.
   • B
     $\frac{v_0^2\sin \theta }{2g}$

     Why not B: Forgot to square $\sin \theta$.
   • C
     $\frac{v_0^2{\sin }^2\theta }{2g}$Correct
   • D
     $\frac{v_0^2{\cos }^2\theta }{2g}$

     Why not D: Used the horizontal component instead of the vertical.
   Explanation

   Vertical initial velocity: $v_y=v_0\sin \theta$. At maximum height $v_y=0$. Kinematics: $0=v_0^2{\sin }^2\theta -2gh$, so $h=\frac{v_0^2{\sin }^2\theta }{2g}$.

   Key takeaway

   Maximum height uses only the vertical component $v_{0y} = v_0\sin\theta$. Set $v_y = 0$ and solve.

7. Question 7 · Medium

   The velocity of a particle is $v(t)=e^{-t}$ m/s for $t\ge 0$. What is the total distance traveled from $t=0$ to $t\to \infty$?

   • A
     $0$ m

     Why not A: Confused displacement (which approaches $+1$ m) with zero net motion.
   • B
     $\frac{1}{e}$ m

     Why not B: Evaluated $v(1)$ rather than the integral.
   • C
     $1$ mCorrect
   • D
     $\infty$ m

     Why not D: Assumed perpetual motion implies infinite distance.
   Explanation

   Since $v(t)=e^{-t}>0$ for all $t\ge 0$, distance $={\int }_0^{\infty }{e}^{-t}dt=[-e^{-t}{]}_0^{\infty }=(0)-(-1)=1$ m.

   Key takeaway

   Total distance $= \int|v|\,dt$. This improper integral converges to $1$ m even though the particle never fully stops.

8. Question 8 · Medium

   A particle has position $\mathbf{r}(t)=t^2\hat{i}+2t^3\hat{j}$ m. What is $|\mathbf{a}|$ at $t=1$ s?

   • A
     $12$ m/s²

     Why not A: Took only the $\hat{j}$ component of acceleration.
   • B
     $14$ m/s²

     Why not B: Added $|a_x|+|a_y|$ instead of taking the vector magnitude.
   • C
     $2\sqrt{37}$ m/s²Correct
   • D
     $\sqrt{40}$ m/s²

     Why not D: Used velocity components $v_x=2,v_y=6$ instead of acceleration components.
   Explanation

   $\mathbf{v}(t)=2t\hat{i}+6t^2\hat{j}$. $\mathbf{a}(t)=2\hat{i}+12t\hat{j}$. At $t=1$: $\mathbf{a}=2\hat{i}+12\hat{j}$, so $|\mathbf{a}|=\sqrt{4+144}=\sqrt{148}=2\sqrt{37}$ m/s².

   Key takeaway

   Differentiate each component of $\mathbf{r}(t)$ twice for $\mathbf{a}(t)$, then use $|\mathbf{a}| = \sqrt{a_x^2 + a_y^2}$.

9. Question 9 · Medium

   A car decelerates uniformly from $30$ m/s to rest in $5$ s. What is the magnitude of deceleration and the stopping distance?

   • A
     $6$ m/s², $75$ mCorrect
   • B
     $5$ m/s², $75$ m

     Why not B: Used $a=30/6$ (arithmetical error).
   • C
     $6$ m/s², $150$ m

     Why not C: Used $d=v_{0}t$ without the $\frac{1}{2}$ factor.
   • D
     $6$ m/s², $90$ m

     Why not D: Computed average velocity as $18$ m/s instead of $15$ m/s.
   Explanation

   $|a|=\frac{|\Delta v|}{\Delta t}=\frac{30}{5}=6$ m/s². Stopping distance: $d=v_{0}t+\frac{1}{2}(-6)(25)=150-75=75$ m. Equivalently, $d=\bar{v}t=15\times 5=75$ m.

   Key takeaway

   For uniform deceleration, $|a| = \Delta v/\Delta t$ and $d = (v_0 + v_f)t/2$.

10. Question 10 · Hard

    A particle's position is $x(t)=A\sin (\omega t)$. Using calculus, find $a(t)$ and identify the relationship between $a$ and $x$.

    • A
      $a(t)=A\omega \cos (\omega t)$; $a=\omega x$

      Why not A: Differentiated only once and formed a wrong proportionality.
    • B
      $a(t)=-A\omega \sin (\omega t)$; $a=-x$

      Why not B: Lost one factor of $\omega$ in the second derivative.
    • C
      $a(t)=-A{\omega }^2\sin (\omega t)$; $a=-{\omega }^{2}x$Correct
    • D
      $a(t)=A{\omega }^2\cos (\omega t)$; $a={\omega }^{2}x$

      Why not D: Incorrect trig function and sign after double differentiation.
    Explanation

    $v(t)=\frac{dx}{dt}=A\omega \cos (\omega t)$. $a(t)=\frac{dv}{dt}=-A{\omega }^2\sin (\omega t)=-{\omega }^{2}x(t)$. The result $a=-{\omega }^{2}x$ is the defining equation of simple harmonic motion.

    Key takeaway

    Differentiating $\sin(\omega t)$ twice brings a factor $\omega^2$ and a sign flip. The relationship $a = -\omega^2 x$ identifies SHM.

11. Question 11 · Hard

    The velocity of a particle is $v(t)=t^2-4t+3$ m/s. What is the total distance traveled from $t=0$ to $t=4$ s?

    (Hint: factor $v$ and identify where it changes sign.)

    • A
      $\frac{4}{3}$ m

      Why not A: Computed net displacement ${\int }_0^{4}vdt$ which gives $\frac{4}{3}$ m.
    • B
      $4$ mCorrect
    • C
      $\frac{16}{3}$ m

      Why not C: Integrated $|v|$ on $[0,3]$ only and stopped.
    • D
      $6$ m

      Why not D: Evaluated $|v(0)|+|v(4)|$ rather than integrating.
    Explanation

    $v(t)=(t-1)(t-3)$. Sign changes at $t=1$ and $t=3$: positive on $[0,1)$, negative on $(1,3)$, positive on $(3,4]$.

    ${\int }_0^{1}vdt={\left[\frac{t^3}{3}-2t^2+3t\right]}_0^1=\frac{1}{3}-2+3=\frac{4}{3}$.

    ${\int }_1^{3}vdt={\left[\frac{t^3}{3}-2t^2+3t\right]}_1^3=(9-18+9)-(\frac{1}{3}-2+3)=0-\frac{4}{3}=-\frac{4}{3}$, distance $=\frac{4}{3}$.

    ${\int }_3^{4}vdt={\left[\frac{t^3}{3}-2t^2+3t\right]}_3^4=(\frac{64}{3}-32+12)-(9-18+9)=\frac{4}{3}-0=\frac{4}{3}$.

    Total distance $=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4$ m.

    Key takeaway

    Total distance $= \int|v(t)|\,dt$. Split at every zero of $v(t)$ and sum absolute values of each segment's integral.

12. Question 12 · Hard

    A particle's position components are $x(t)=v_{0x}t$ and $y(t)=v_{0y}t-\frac{1}{2}g{t}^2$. Which expression correctly gives $y$ as a function of $x$ (the trajectory equation)?

    • A
      $y=\frac{v_{0y}}{v_{0x}}x-\frac{g}{v_{0x}^2}{x}^2$

      Why not A: Forgot the factor of $\frac{1}{2}$ when substituting $t=x/v_{0x}$ into the $\frac{1}{2}g{t}^2$ term.
    • B
      $y=\frac{v_{0y}}{v_{0x}}x-\frac{g}{2v_{0x}^2}{x}^2$Correct
    • C
      $y=\frac{v_{0x}}{v_{0y}}x-\frac{g}{2v_{0y}^2}{x}^2$

      Why not C: Swapped $v_{0x}$ and $v_{0y}$ when solving $t=x/v_{0x}$.
    • D
      $y=\frac{v_{0y}}{v_{0x}}x+\frac{g}{2v_{0x}^2}{x}^2$

      Why not D: Sign error: gravity reduces height, so the $x^2$ term must be subtracted.
    Explanation

    From $x=v_{0x}t$: $t=\frac{x}{v_{0x}}$. Substitute:
    $$y=v_{0y}\cdot \frac{x}{v_{0x}}-\frac{1}{2}g{\left(\frac{x}{v_{0x}}\right)}^2=\frac{v_{0y}}{v_{0x}}x-\frac{g}{2v_{0x}^2}{x}^2.$$
    In terms of launch angle $\theta$: $v_{0y}/v_{0x}=\tan \theta$ and $v_{0x}=v_0\cos \theta$, giving the standard form $y=x\tan \theta -\frac{g}{2v_0^2{\cos }^2\theta }{x}^2$.

    Key takeaway

    Eliminate $t$ from parametric equations by solving $x(t)$ for $t$ and substituting into $y(t)$. The result is a downward parabola — the defining shape of projectile trajectories.