CLASSICAL-MECHANICS

Beyond Constant Acceleration: Polynomial and Piecewise Integration

IKey Formulas

Formula	Name	Description
$a(t) = b_0 + b_1 t + b_2 t^2$	Polynomial acceleration	Constant baseline, linear ramp, quadratic curvature
$v(t) = v_0 + b_0 t + \frac{b_1}{2}t^2 + \frac{b_2}{3}t^3$	Velocity	Integrate $a(t)$ term by term
$x(t) = x_0 + v_0 t + \frac{b_0}{2}t^2 + \frac{b_1}{6}t^3 + \frac{b_2}{12}t^4$	Position	Integrate $v(t)$ term by term
$\int_{t_1}^{t} b(t'-t_1)\,dt' = \frac{b}{2}(t-t_1)^2$	Piecewise braking	Integrate from the interval's start $t_1$ , match boundaries

IIInteractive Visualizations

Visualization 1 — Integration raises the degree

Nudge a coefficient of $a(t)$ and watch it climb into $v(t)$ and $x(t)$ , one power of $t$ heavier each step.

b₀ (baseline) 2 b₁ (ramp) 1 b₂ (curvature) 0.5

💡 Each integration multiplies a term's running denominator by its new exponent — the $b_1$ chain runs $1 \to 2 \to 6$ and the $b_2$ chain $1 \to 3 \to 12$ .

Visualization 2 — Velocity is the area under a(t)

The velocity gained by time $t$ is exactly the area swept out under the acceleration curve from $0$ to $t$ .

Upper limit t (s) 2

Visualization 3 — Stitching a piecewise trip together

One bicycle speed lets it reach the braking car exactly as the car comes to rest — which value of $v_b$ ?

Bicycle speed v_b (m/s) 12

💡 The car's curve only bends after $t_1 = 1\text{ s}$ : during the coast $a = 0$ , and the boundary value $x_c(t_1) = 12\text{ m}$ carries forward as the braking phase's initial condition.

IIIQuiz Questions

Question 1

An object has acceleration $a(t) = b_0 + b_1 t + b_2 t^2$ and starts with velocity $v_0$ . Integrating once, what is the velocity $v(t)$ ?

✅ Correct! Each term gains one power of t and divides by the new exponent.

❌ Not quite. Use the power rule term by term: the t² term divides by 2 and the t³ term divides by 3.

Show solution

Solution:

Integrate each term of $a(t)$ from $0$ to $t$ and add back $v_0$ . The power rule raises each exponent by one and divides by the new exponent:

$v(t) = v_0 + \int_0^t (b_0 + b_1 t' + b_2 t'^2)\,dt' = v_0 + b_0 t + \frac{b_1}{2}t^2 + \frac{b_2}{3}t^3$

Each acceleration coefficient produces a velocity term one degree higher.

Question 2

Integrating the velocity $v(t) = v_0 + b_0 t + \frac{b_1}{2}t^2 + \frac{b_2}{3}t^3$ to get position, what denominator appears on the $b_1$ term in $x(t)$ ?

✅ Correct! The existing 2 times the new exponent 3 gives 6.

❌ Not quite. Multiply the denominator already on the term (2) by the new exponent (3).

Show solution

Solution:

In $v(t)$ the $b_1$ term is $\frac{b_1}{2}t^2$ . Integrating raises $t^2$ to $t^3$ and divides by the new exponent 3:

$\int_0^t \frac{b_1}{2}t'^2\,dt' = \frac{b_1}{2}\cdot\frac{t^3}{3} = \frac{b_1}{6}t^3$

The denominator is $2 \times 3 = 6$ , giving the position term $\frac{b_1}{6}t^3$ .

Question 3

True or False: For piecewise acceleration, when you integrate the braking phase you should always integrate starting from $t = 0$ , the very beginning of the motion.

✅ Correct! Integrate each interval from its own start and match boundary values.

❌ Not quite. The braking phase is integrated from t₁, with v and x carried forward from the previous interval.

Show solution

Solution:

False. For piecewise acceleration, you integrate each interval separately, starting from the beginning of that interval — here the braking phase is integrated from $t_1$ , not from $0$ :

$v_c(t) - v_c(t_1) = \int_{t_1}^{t} b(t' - t_1)\,dt' = \frac{b}{2}(t - t_1)^2$

Velocity and position must be continuous at the boundary, so $v_c(t_1) = 12\text{ m/s}$ and $x_c(t_1) = 12\text{ m}$ carry forward as the initial conditions for the braking phase. Resetting these to zero is the most common error in piecewise problems.

Question 4

In the bicycle-and-car problem, the car stops at $t_2 = 3\text{ s}$ at $x_c = 28\text{ m}$ . The bicycle starts at $-17\text{ m}$ at constant speed $v_b$ and meets the car exactly then. What is $v_b$ ?

✅ Correct! From −17 + 3v_b = 28 you get v_b = 15 m/s.

❌ Not quite. Set −17 + 3v_b = 28 and solve: it covers 45 m of ground in 3 s.

Show solution

Solution:

The bicycle moves at constant speed, so its position is $x_b(t) = -17 + v_b t$ . Set it equal to the car's meeting position at $t = 3\text{ s}$ :

$-17 + 3 v_b = 28$ $3 v_b = 45$ $v_b = 15 \text{ m/s}$

Setting the two position equations equal at the encounter time and solving for the unknown is the general recipe for meeting problems.

Solved: 0 / 4