AP Physics 1 Kinematics — Worked Answer Explanations

Unit 1 · 12% of the AP exam · 10 questions explained

1. Question 1 · Easy

   A car traveling at $20\text{m/s}$ brakes uniformly and stops in $4.0\text{s}$. What is the magnitude of its acceleration?

   • A
     $2.5{\text{m/s}}^2$

     Why not A: Half-counted the velocity change, perhaps treating $\Delta v=10\text{m/s}$.
   • B
     $5.0{\text{m/s}}^2$Correct
   • C
     $8.0{\text{m/s}}^2$

     Why not C: Divided velocity by an incorrect time interval.
   • D
     $80{\text{m/s}}^2$

     Why not D: Multiplied instead of divided, $20\times 4=80$.
   Explanation

   Use $a=\frac{\Delta v}{\Delta t}=\frac{0-20}{4.0}=-5.0{\text{m/s}}^2$. The magnitude is $5.0{\text{m/s}}^2$.

   Key takeaway

   Average acceleration is the change in velocity divided by elapsed time.

2. Question 2 · Easy

   The velocity-vs-time graph of a moving object is a straight line that starts at $v=-4\text{m/s}$ at $t=0$ and reaches $v=+6\text{m/s}$ at $t=5\text{s}$. What is the average acceleration over this interval?

   • A
     $0.4{\text{m/s}}^2$

     Why not A: Subtracted incorrectly: used $|6|-|-4|=2$, then $2/5$.
   • B
     $1.2{\text{m/s}}^2$

     Why not B: Took $|-4|+6=10$ but divided by an incorrect time.
   • C
     $2.0{\text{m/s}}^2$Correct
   • D
     $10{\text{m/s}}^2$

     Why not D: Forgot to divide by the elapsed time.
   Explanation

   $a_{avg}=\frac{v_f-v_i}{\Delta t}=\frac{6-(-4)}{5}=\frac{10}{5}=2.0{\text{m/s}}^2$. The negative initial velocity is part of the change.

   Key takeaway

   When velocity changes sign, $\Delta v$ adds the magnitudes — be careful with signs.

3. Question 3 · Medium

   A ball is thrown straight up with an initial speed of $15\text{m/s}$. Taking $g=10{\text{m/s}}^2$ and ignoring air resistance, what is the maximum height reached above the launch point?

   • A
     $7.5\text{m}$

     Why not A: Used $h=v_{0}t/2$ with $t=1.0\text{s}$, missing the proper kinematic relation.
   • B
     $11.25\text{m}$Correct
   • C
     $15\text{m}$

     Why not C: Confused the launch speed with a distance value.
   • D
     $22.5\text{m}$

     Why not D: Doubled the correct answer, perhaps adding rise and fall together.
   Explanation

   At maximum height, $v_f=0$. Using $v_f^2=v_0^2-2gh$, we get $0=(15{)}^2-2(10)h$, so $h=\frac{225}{20}=11.25\text{m}$.

   Key takeaway

   At the apex of a vertical throw, the velocity is zero — apply $v_f^2 = v_0^2 - 2gh$.

4. Question 4 · Medium

   An object moves along the x-axis with the position function $x(t)=3t^2-12t+5$ (meters, with $t$ in seconds). At what time does the object momentarily come to rest?

   • A
     $t=0\text{s}$

     Why not A: Confused initial position with zero velocity; at $t=0$ the object has nonzero velocity.
   • B
     $t=2\text{s}$Correct
   • C
     $t=4\text{s}$

     Why not C: Set $x(t)=0$ rather than $v(t)=0$.
   • D
     The object never comes to rest.

     Why not D: Missed that the velocity polynomial has a real root.
   Explanation

   Velocity is $v(t)=\frac{dx}{dt}=6t-12$. Setting $v(t)=0$ gives $t=2\text{s}$. (Although AP Physics 1 is algebra-based, this can also be read off the slope of $x(t)$ being zero at the vertex of the parabola.)

   Key takeaway

   An object is momentarily at rest when its instantaneous velocity is zero, not when its position is zero.

5. Question 5 · Medium

   Two students start from the same point. Student A runs at a constant $4.0\text{m/s}$. Student B starts from rest at the same instant and accelerates uniformly at $2.0{\text{m/s}}^2$. How long after the start does Student B catch up with Student A?

   • A
     $2.0\text{s}$

     Why not A: Equated the speeds rather than the positions.
   • B
     $4.0\text{s}$Correct
   • C
     $6.0\text{s}$

     Why not C: Off by a factor in solving the quadratic.
   • D
     Student B never catches up.

     Why not D: Missed that constant acceleration eventually overtakes constant velocity.
   Explanation

   Set positions equal: $4t=\frac{1}{2}(2)t^2$, so $4t=t^2$. Dividing by $t$ (excluding $t=0$): $t=4.0\text{s}$.

   Key takeaway

   Two objects are at the same location when their position functions are equal — set them equal and solve.

6. Question 6 · Medium

   A projectile is launched from ground level at a speed of $30\text{m/s}$ at an angle of $30°$ above the horizontal. Taking $g=10{\text{m/s}}^2$ and ignoring air resistance, how long is the projectile in the air before returning to the ground?

   • A
     $1.5\text{s}$

     Why not A: Time to apex only — forgot the symmetric descent.
   • B
     $3.0\text{s}$Correct
   • C
     $5.2\text{s}$

     Why not C: Used the horizontal component of velocity instead of vertical.
   • D
     $6.0\text{s}$

     Why not D: Used full launch speed instead of vertical component.
   Explanation

   Vertical component: $v_{0y}=30\sin 30°=15\text{m/s}$. Time to apex: $t_{up}=\frac{v_{0y}}{g}=1.5\text{s}$. By symmetry, total time of flight is $2t_{up}=3.0\text{s}$.

   Key takeaway

   For a projectile launched and landing at the same height, total flight time is $2 v_{0y}/g$.

7. Question 7 · Medium

   A stone is dropped from a tall cliff. Ignoring air resistance and taking $g=10{\text{m/s}}^2$, how far does it fall during the third second of its fall (between $t=2\text{s}$ and $t=3\text{s}$)?

   • A
     $10\text{m}$

     Why not A: Used the speed gained in 1 s as if it were a distance.
   • B
     $20\text{m}$

     Why not B: Treated the average speed during the interval as the speed at the start.
   • C
     $25\text{m}$Correct
   • D
     $45\text{m}$

     Why not D: Reported total distance fallen by $t=3\text{s}$ rather than the distance during the third second.
   Explanation

   Distance fallen from rest in time $t$ is $\frac{1}{2}g{t}^2$. From $t=0$ to $3\text{s}$: $\frac{1}{2}(10)(9)=45\text{m}$. From $t=0$ to $2\text{s}$: $\frac{1}{2}(10)(4)=20\text{m}$. Difference: $25\text{m}$.

   Key takeaway

   Distance during the $n$th second of free fall from rest is $\tfrac{1}{2}g(n^2 - (n-1)^2)$ — they grow linearly with time.

8. Question 8 · Medium

   A car accelerates uniformly from rest. After traveling $20\text{m}$, its speed is $10\text{m/s}$. What is its speed after it has traveled an additional $60\text{m}$ (i.e., $80\text{m}$ total)?

   • A
     $15\text{m/s}$

     Why not A: Linearly scaled speed with distance.
   • B
     $20\text{m/s}$Correct
   • C
     $30\text{m/s}$

     Why not C: Doubled the speed for tripled additional distance.
   • D
     $40\text{m/s}$

     Why not D: Quadrupled speed proportional to total distance ratio without taking square root.
   Explanation

   From $v^2=v_0^2+2a\Delta x$ at $20\text{m}$: $(10{)}^2=0+2a(20)$, so $a=2.5{\text{m/s}}^2$. At $80\text{m}$: $v^2=0+2(2.5)(80)=400$, so $v=20\text{m/s}$.

   Key takeaway

   Under constant acceleration from rest, speed scales with the square root of distance.

9. Question 9 · Medium

   An object's position as a function of time is shown in a graph (not reproduced here): position is constant at $x=4\text{m}$ for $0\le t<2\text{s}$, then increases linearly to $x=10\text{m}$ at $t=5\text{s}$, then decreases linearly back to $x=4\text{m}$ at $t=8\text{s}$. What is the object's average velocity over the interval from $t=0$ to $t=8\text{s}$?

   • A
     $0\text{m/s}$Correct
   • B
     $0.75\text{m/s}$

     Why not B: Used average speed (total path / total time) rather than displacement / time.
   • C
     $1.25\text{m/s}$

     Why not C: Used the slope of one of the segments only.
   • D
     $2.0\text{m/s}$

     Why not D: Used the maximum slope on the graph.
   Explanation

   Average velocity is total displacement divided by total time. The object starts and ends at $x=4\text{m}$, so $\Delta x=0$ and $\bar{v}=0$. Average speed (not asked) would be $12\text{m}/8\text{s}=1.5\text{m/s}$.

   Key takeaway

   Average velocity uses displacement (final − initial), not total distance traveled.

10. Question 10 · Hard

    Two students compare their commutes. Student X drives the first half of the distance at $30\text{m/s}$ and the second half at $60\text{m/s}$. Student Y drives the first half of the time at $30\text{m/s}$ and the second half of the time at $60\text{m/s}$. Which student has the larger average speed for the trip?

    • A
      Student X, because they spend less time at the lower speed.

      Why not A: Misidentifies which split format gives more time at the slow speed.
    • B
      Student Y, because the average is just the simple mean of the two speeds.Correct
    • C
      They have the same average speed, $45\text{m/s}$.

      Why not C: Assumes 'average speed' is always the simple mean regardless of how the trip is split.
    • D
      Cannot be determined without more information.

      Why not D: All needed information is given.
    Explanation

    Student Y's average is the simple mean: $(30+60)/2=45\text{m/s}$. Student X's average is the harmonic mean: $\bar{v}=\frac{2v_1{v}_2}{v_1+v_2}=\frac{2(30)(60)}{90}=40\text{m/s}$. Splitting by distance puts more time at the slow speed, dragging the average down.

    Key takeaway

    Equal-distance splits give the harmonic mean; equal-time splits give the arithmetic mean. The arithmetic mean is always larger.