Rectilinear Kinematics
Velocity from Position-Dependent Acceleration — Numerical Integration
A particle moves along a straight line with an acceleration defined as a function of position:
a = 5 / (3s1/3 + s5/2) m/s²,
where s is in meters. The particle starts from rest at s = 1 m.
Determine the velocity of the particle when it reaches s = 2 m,
using a numerical method to evaluate the resulting integral.
What We Know
- Acceleration: a(s) = 5 / (3s1/3 + s5/2) m/s²
- Initial condition: v = 0 (starts from rest) at s = 1 m
- Find: velocity v when s = 2 m
- Method: numerical integration (Simpson’s Rule)
Why v dv = a ds? The acceleration is given as a function of position s, not time t. Neither integrating a dt (which requires a as a function of t) nor the constant-acceleration formulas apply here. The chain rule gives us the bridge: a = dv/dt = (dv/ds)(ds/dt) = v(dv/ds), so v dv = a ds. This separates velocity and position onto opposite sides, allowing integration from s = 1 m to s = 2 m. The right-hand integral has no closed form, so Simpson’s Rule gives the numerical answer.
Part A — Setting Up the v dv = a ds Integral
Step 1
Derive the Working Equation Using the Chain Rule
Since a is a function of s, we eliminate time using the identity a = v dv/ds:
\[ v\, dv = a\, ds = \frac{5}{3s^{1/3} + s^{5/2}}\, ds \]
Integrate both sides — left side from v = 0 to v, right side from s = 1 m to s = 2 m:
\[ \int_{0}^{v} v\, dv = 5\int_{1}^{2} \frac{ds}{3s^{1/3} + s^{5/2}} \]
Evaluating the left side analytically:
\[ \left[\frac{v^2}{2}\right]_{0}^{v} = \frac{v^2}{2} \]
Define f(s) for the right-hand integrand:
\[ f(s) = \frac{1}{3s^{1/3} + s^{5/2}} \]
\[ \frac{v^2}{2} = 5\int_{1}^{2} f(s)\, ds \qquad \cdots (1) \]
Why can’t we integrate analytically? The denominator 3s1/3 + s5/2 mixes a cube-root term and a fractional-power term — no standard antiderivative exists for this combination. Numerical integration is the prescribed and correct approach.
Part B — Numerical Integration via Simpson’s Rule
Step 2
Apply Simpson’s Rule with Three Points
Simpson’s Rule for a single interval [a, b] uses three equally-spaced evaluation points and provides a parabolic approximation to the integrand:
\[ \int_{a}^{b} f(s)\, ds \approx \frac{b-a}{6}\left[f(a) + 4f\!\left(\tfrac{a+b}{2}\right) + f(b)\right] \]
Here a = 1 m, b = 2 m, and the midpoint is (1+2)/2 = 1.5 m. Evaluate f(s) at each point:
| Point | s (m) | 3s1/3 | s5/2 | Denominator | f(s) | Weight |
|---|---|---|---|---|---|---|
| f(a) | 1.000 | 3.0000 | 1.0000 | 4.0000 | 0.2500 | 1 |
| f(mid) | 1.500 | 3.4341 | 2.7557 | 6.1898 | 0.1616 | 4 |
| f(b) | 2.000 | 3.7798 | 5.6569 | 9.4367 | 0.1060 | 1 |
Applying the formula:
\[
\int_{1}^{2} f(s)\, ds \approx \frac{2-1}{6}\left[0.2500 + 4(0.1616) + 0.1060\right]
\]
\[
= \frac{1}{6}\left[0.2500 + 0.6464 + 0.1060\right] = \frac{1.0024}{6}
\]
\[ \int_{1}^{2} f(s)\, ds \approx 0.1671 \]
How accurate is Simpson’s Rule? Simpson’s Rule fits a parabola through three points, giving exact results for polynomials up to degree 3. For smooth non-polynomial functions like this one, a single-interval application gives a good approximation — and the problem explicitly permits this numerical approach.
Part C — Solve for Velocity
Step 3
Substitute Back and Solve for v
Substituting the numerical result into equation (1):
\[ \frac{v^2}{2} = 5 \times 0.1671 = 0.8353 \]
\[ v^2 = 2 \times 0.8353 = 1.6706 \]
\[ v = \sqrt{1.6706} \]
\[ v = 1.29 \ \text{m/s} \]
Physical interpretation: The particle starts from rest at s = 1 m and, under the influence of the position-dependent acceleration, reaches a velocity of 1.29 m/s by the time it arrives at s = 2 m. Note that the acceleration decreases as s increases (larger denominator → smaller a), so the particle gains velocity more slowly as it moves further from the origin — which is consistent with the relatively modest final speed despite traveling 1 m.
Final Answer
VELOCITY AT s = 2 m
v = 1.29 m/s
METHOD USED
v dv = a ds + Simpson’s Rule (numerical integration)
INTEGRAL VALUE
∫ f(s) ds from 1 to 2 ≈ 0.1671 (Simpson’s Rule, 3 points)