Rectilinear Kinematics
Velocity as a Function of Position under Constant Acceleration
A particle moves along a straight line with a constant (but unknown) acceleration.
Two states of the particle are observed: when s = 4 ft, the velocity is v = 3 ft/s,
and when s = 10 ft, the velocity has grown to v = 8 ft/s.
Using these two data points, determine the velocity of the particle as a general
function of its position s.
What We Know
- At position s₀ = 4 ft → velocity v₀ = 3 ft/s
- At position s = 10 ft → velocity v = 8 ft/s
- Acceleration is constant (magnitude unknown)
- Goal: express v as a function of s for any position
Why v² = v₀² + 2a(s − s₀)? This kinematic equation links velocity and position directly without involving time — exactly what the problem asks for. With two known (s, v) pairs and constant acceleration, we first solve for a, then rewrite the same equation leaving s and v as variables to get the general function v(s).
For any particle undergoing constant acceleration, the relationship between velocity and position is:
\[ v^2 = v_0^2 + 2a(s – s_0) \]
This equation is derived by eliminating time from the two fundamental kinematic equations. It is especially useful here because the problem gives us two position–velocity pairs rather than time information.
Why this equation? The problem never mentions time — it only gives two snapshots of (position, velocity). The v²–s equation is the only standard kinematic equation that connects these two quantities without needing t.
Plug in both known data points — s₀ = 4 ft, v₀ = 3 ft/s, s = 10 ft, v = 8 ft/s — and solve for a:
\[ (8)^2 = (3)^2 + 2a(10 – 4) \]
\[ 64 = 9 + 12a \]
\[ 12a = 55 \]
\[ a = \frac{55}{12} \approx 4.583 \ \text{ft/s}^2 \]
Exact vs. decimal: The exact value is 55/12 ft/s². Using 4.583 in subsequent steps introduces a small rounding error. We carry the exact fraction through the algebra for a cleaner final expression.
Now substitute the known reference point (s₀ = 4 ft, v₀ = 3 ft/s) and the computed acceleration a = 55/12 ft/s² back into the same equation, but this time leave s and v as variables:
\[ v^2 = (3)^2 + 2\left(\frac{55}{12}\right)(s – 4) \]
\[ v^2 = 9 + \frac{55}{6}(s – 4) \]
\[ v^2 = 9 + 9.17(s – 4) \quad \text{(decimal form)} \]
Taking the positive square root (particle moves in the positive direction):
\[ v = \sqrt{9 + 9.17(s – 4)} \quad \text{ft/s} \]
Verification: At s = 4 ft: v = √(9 + 0) = 3 ft/s ✓ | At s = 10 ft: v = √(9 + 9.17 × 6) = √(9 + 55) = √64 = 8 ft/s ✓ — both original data points are exactly reproduced.
The final expression gives velocity at any position s along the path (provided s ≥ 4 ft, where the particle starts):
\[ v(s) = \sqrt{9 + 9.17(s – 4)} \quad \text{ft/s} \]
A few quick values to build physical intuition:
\[
\begin{array}{c|c}
s \ (\text{ft}) & v \ (\text{ft/s}) \\
\hline
4 & 3.00 \\
7 & 5.95 \\
10 & 8.00 \\
15 & 10.54 \\
\end{array}
\]
Physical meaning: Velocity increases with position because the particle is accelerating in the positive direction. The square-root shape means velocity grows quickly at first and more gradually as s increases — a hallmark of constant-acceleration motion when plotted against position rather than time.
Final Answers
CONSTANT ACCELERATION
a = 55/12 ≈ 4.583 ft/s²
VELOCITY AS A FUNCTION OF POSITION
v = √[9 + 9.17(s − 4)] ft/s
