Seven problems that exercise every tool from Chapter 2. Each begins with a scenario, then walks through the full solution — emphasizing which §2.x section provides the tool and why that formula is chosen. If a particular tool feels shaky, click through to the underlying section.
Problem 1 — a picnic trip (§2.1)
Scenario: a cyclist leaves home and rides north. In the first 30 minutes she covers 6 km. Then a 10-minute rest. Then 4 km back and she rests in a park. Compute:
a) total displacement
b) total distance
c) average velocity
d) average speed
Solution:
Axis: north positive. Origin = home. Total time: assume the return leg took 20 min, so total = 30 + 10 + 20 = 60 min = 1 h.
Displacement: only endpoints matter:
$$
\Delta x = x_\text{park} – x_\text{home} = (6 – 4) – 0 = +2\ \mathrm{km}
$$
Distance: sum of leg lengths:
$$
d = 6 + 4 = 10\ \mathrm{km}
$$
(Rest doesn’t add distance since she’s stationary.)
$$
v_\text{avg} = \frac{\Delta x}{\Delta t} = \frac{2\ \mathrm{km}}{1\ \mathrm{h}} = 2\ \mathrm{km/h}
$$
$$
s_\text{avg} = \frac{d}{\Delta t} = \frac{10\ \mathrm{km}}{1\ \mathrm{h}} = 10\ \mathrm{km/h}
$$
Insight: average velocity 2, average speed 10 — a factor of 5. Because most of the trip was back-and-forth. Distance and displacement can differ dramatically.
Problem 2 — instantaneous velocity from a function (§2.2)
Scenario: a particle moves along the x axis with
$$
x(t) = 4 t^3 – 6 t^2 + 2 t\quad (\mathrm{m},\ t\ \text{in seconds})
$$
a) What is its velocity at t = 0.5 s?
b) When is it momentarily at rest?
c) What is its acceleration at those instants?
Solution:
Differentiate for velocity:
$$
v(t) = \frac{dx}{dt} = 12 t^2 – 12 t + 2
$$
a) at t = 0.5:
$$
v(0.5) = 12 \cdot 0.25 – 12 \cdot 0.5 + 2 = 3 – 6 + 2 = -1\ \mathrm{m/s}
$$
Negative — going backward.
b) Rest instants: v = 0.
$$
12 t^2 – 12 t + 2 = 0 \Rightarrow 6 t^2 – 6 t + 1 = 0
$$
$$
t = \frac{6 \pm \sqrt{36 – 24}}{12} = \frac{6 \pm 2\sqrt{3}}{12}
$$
$$
t_1 = \frac{6 – 2\sqrt{3}}{12} \approx 0.211\ \mathrm{s},\quad t_2 = \frac{6 + 2\sqrt{3}}{12} \approx 0.789\ \mathrm{s}
$$
Two turning points — one when the motion reverses, one when it reverses back.
c) Acceleration:
$$
a(t) = \frac{dv}{dt} = 24 t – 12
$$
At t_1 ≈ 0.211: a ≈ 5.07 - 12 = -6.93 m/s² (reversing to backward)
At t_2 ≈ 0.789: a ≈ 18.93 - 12 = +6.93 m/s² (reversing to forward)
Insight: acceleration is not zero at the moments of rest. If it were, the particle would stay stuck. §2.3 established this.
Problem 3 — stopping distance at two speeds (§2.4)
Scenario: a new driver wants to know: if she can stop in 12 m at 50 km/h, how much distance does she need at 100 km/h (twice as fast)? Assume the same braking capability (same acceleration magnitude).
Solution:
From §2.4 equation 3 (no time): v² = v_0² + 2a(x - x_0). With v = 0 (stopping):
$$
x_\text{stop} = \frac{v_0^2}{2|a|}
$$
Key: distance scales with the square of velocity. So:
$$
\frac{x_\text{stop}(100)}{x_\text{stop}(50)} = \left(\frac{100}{50}\right)^2 = 4
$$
$$
x_\text{stop}(100) = 4 \cdot 12 = 48\ \mathrm{m}
$$
Insight: double the speed = four times the stopping distance. If a driver on a foggy road at 100 km/h sees a stopped car 40 m ahead, there’s no time to react — even with hand on the brake. That’s a physics fact, not a driving choice.
Problem 4 — throwing a ball to the ceiling (§2.5)
Scenario: you throw a ball from height y_0 = 1 m above the floor straight up at v_0 = 8 m/s. The ceiling is at y = 3.5 m.
a) Does the ball reach the ceiling?
b) If yes, what is its speed when it hits the ceiling?
c) When does it reach the ceiling?
Solution:
Axis: up positive. a = -g = -9.8 m/s².
a) Check if it reaches:
First compute the maximum possible height. At the peak, v = 0. From equation 3:
$$
0 = 8^2 – 2 \cdot 9.8 \cdot (y_\text{max} – 1)
$$
$$
y_\text{max} = 1 + \frac{64}{19.6} \approx 1 + 3.27 = 4.27\ \mathrm{m}
$$
4.27 > 3.5 — yes, it reaches. With about 0.77 m of margin.
b) Speed at ceiling height: from equation 3 with y = 3.5:
$$
v^2 = 8^2 – 2 \cdot 9.8 \cdot (3.5 – 1) = 64 – 49 = 15
$$
$$
v = \pm\sqrt{15} \approx \pm 3.87\ \mathrm{m/s}
$$
The ball is still on its way up (first crossing of the ceiling), so v = +3.87 m/s.
c) Time to ceiling: from equation 1:
$$
3.87 = 8 – 9.8 t \Rightarrow t = \frac{8 – 3.87}{9.8} \approx 0.42\ \mathrm{s}
$$
Insight: the ball reaches the ceiling in less than half a second — faster than intuition suggests. And it tries to go higher, so it will hit the ceiling. A taller ceiling (say 5 m) would give a “no” answer.
Problem 5 — a train at variable speed (§2.6)
Scenario: a train leaves station A. Its v-t diagram shows:
[0, 60]s: constant acceleration — velocity from0to30 m/s[60, 240]s: constant velocity30 m/s(cruise)[240, 300]s: linear braking — velocity to0
What is the distance from station A to B?
Solution:
From §2.6: area under v-t = displacement. Compute the three pieces:
Piece 1 (accel): triangle, base 60 s, height 30 m/s:
$$
A_1 = \frac{1}{2} \cdot 60 \cdot 30 = 900\ \mathrm{m}
$$
Piece 2 (cruise): rectangle, base 180 s, height 30 m/s:
$$
A_2 = 180 \cdot 30 = 5400\ \mathrm{m}
$$
Piece 3 (brake): triangle, base 60 s, height 30 m/s:
$$
A_3 = \frac{1}{2} \cdot 60 \cdot 30 = 900\ \mathrm{m}
$$
Total distance:
$$
d = 900 + 5400 + 900 = 7200\ \mathrm{m} = 7.2\ \mathrm{km}
$$
Insight: we didn’t write a single kinematic equation. Just plot geometry. That’s the power of §2.6.
Problem 6 — two cars on the highway (§2.3 + §2.4)
Scenario: car A cruises at constant 25 m/s on the highway. Car B, 50 m behind, starts accelerating from speed 20 m/s at a = 1.5 m/s². When does B catch A (pass it)?
Solution:
Positive axis: forward. Origin = B’s initial position. Time t = 0 = start of B’s acceleration.
Position of each car:
Car A (constant 25):
$$
x_A(t) = 50 + 25 t
$$
(Because at t=0, A is 50 m ahead of B.)
Car B (accel 1.5 from 20):
$$
x_B(t) = 20 t + \frac{1}{2} \cdot 1.5 \cdot t^2 = 20 t + 0.75 t^2
$$
Catch condition: x_A = x_B:
$$
50 + 25 t = 20 t + 0.75 t^2
$$
$$
0.75 t^2 – 5 t – 50 = 0
$$
Multiply by 4/3 to simplify:
$$
t^2 – \frac{20}{3} t – \frac{200}{3} = 0
$$
Quadratic formula:
$$
t = \frac{20/3 \pm \sqrt{(20/3)^2 + 800/3}}{2}
$$
$$
= \frac{6.67 \pm \sqrt{2800/9}}{2} = \frac{6.67 \pm 17.63}{2}
$$
Positive root:
$$
t \approx 12.2\ \mathrm{s}
$$
Insight: about 12 seconds. At that instant, B’s velocity:
$$
v_B = 20 + 1.5 \cdot 12.2 = 38.3\ \mathrm{m/s}
$$
So B passes A moving 13 m/s faster. Distance covered by B:
$$
x_B = 20 \cdot 12.2 + 0.75 \cdot 148.8 \approx 244 + 112 = 356\ \mathrm{m}
$$
Distance covered by A: 25 \cdot 12.2 = 305 m. Difference: 356 - 305 = 51 m — matching the initial 50 m gap (with rounding). ✓
Problem 7 — integrating with variable acceleration (§2.6)
Scenario: a rocket starts from rest. Its acceleration decays with time (fuel depleting):
$$
a(t) = a_0\, e^{-t/\tau}
$$
with a_0 = 20 m/s² and τ = 10 s (fuel-depletion time constant).
a) What is the rocket’s velocity as t → ∞?
b) Its position at t = 30 s?
Solution:
a) Terminal velocity:
$$
v(t) = \int_0^t a_0 e^{-t’/\tau} dt’ = a_0 \tau \left(1 – e^{-t/\tau}\right)
$$
As t → ∞, e^{-t/τ} → 0:
$$
v_\infty = a_0 \tau = 20 \cdot 10 = 200\ \mathrm{m/s}
$$
Terminal velocity 200 m/s — even with ongoing (decaying) acceleration, velocity approaches a finite limit.
b) Position at t = 30 s:
$$
x(t) = \int_0^t v(t’)\, dt’ = \int_0^t a_0 \tau (1 – e^{-t’/\tau}) dt’
$$
$$
= a_0 \tau \left[t + \tau e^{-t/\tau} – \tau\right]
$$
$$
= a_0 \tau^2 \left[\frac{t}{\tau} – 1 + e^{-t/\tau}\right]
$$
At t = 30, t/τ = 3, e^{-3} ≈ 0.05:
$$
x(30) = 20 \cdot 100 \cdot [3 – 1 + 0.05] = 2000 \cdot 2.05 = 4100\ \mathrm{m}
$$
Insight: in 30 s, the rocket has traveled 4.1 km and reached about 190 m/s (v(30) ≈ 200(1 - 0.05) ≈ 190 m/s). If acceleration had been constant 20 m/s² (no decay), after 30 s the speed would be 600 m/s and the distance 9000 m. Decaying acceleration matters a lot.
Closing note: the 5-step strategy
Solve every problem with these five steps:
- Draw: a quick diagram with axis, positive direction, origin. Even for simple problems.
- Given / find: write what you know and what you want.
- Section: identify which §2.x section provides the tool. Is
aconstant? Is time in the problem? - Equation: pick the one that omits the missing variable (§2.4 table).
- Check: after solving, sanity-check units, magnitude, sign. A
-500 m/sanswer for a moving cyclist is wrong.
Preview of §2.8
Now for practice — problems without step-by-step solutions that you work yourself. Various levels: easy, medium, challenging. §2.8 has ~20 problems that exercise every Chapter 2 tool in fresh scenarios.
📚 See also: Halliday Vol 1, Ch 2 — Worked Examples.
📖 Open reference: OpenStax University Physics Vol 1 — Chapter 3 Solved Examples.
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