Rectilinear Kinematics
Position of a Particle under Constant Deceleration
A particle moves along a straight horizontal path with an initial velocity of
v₀ = 12 ft/s directed to the right, starting from position
s₀ = 0. The particle experiences a constant acceleration of
a = 2 ft/s² acting to the left (opposing the motion).
Determine the position of the particle when t = 10 s.
What We Know
- Initial velocity: v₀ = 12 ft/s (to the right →)
- Initial position: s₀ = 0 ft
- Acceleration magnitude: a = 2 ft/s² (to the left ←)
- Time of interest: t = 10 s
- Sign convention: rightward = positive (+)
Why the constant-acceleration formula? Because the acceleration is a fixed value (not a function of time or position), the standard kinematic equation s = s₀ + v₀t + ½at² applies directly. No integration is needed — the formula is itself the result of integrating constant acceleration twice.
Step 1
Assign the Sign Convention
Before substituting any numbers, we must assign a positive direction consistently. Here we take rightward as positive. This means:
\[
v_0 = +12 \ \text{ft/s}, \qquad
s_0 = 0 \ \text{ft}, \qquad
a = -2 \ \text{ft/s}^2
\]
Why is a negative? The acceleration acts to the left, which is the negative direction under our chosen convention. Using the wrong sign here is the most common error in this type of problem — always check the direction of acceleration against your sign convention before proceeding.
Step 2
Write the Position Equation
For constant acceleration, position as a function of time is given by the standard kinematic equation:
\[ s = s_0 + v_0 t + \frac{1}{2} a t^2 \]
This equation comes from integrating constant acceleration twice with respect to time. The three terms represent: (1) the initial position, (2) displacement due to the initial velocity, and (3) displacement due to the constant acceleration.
Physical note: Notice that the third term is negative here (since a = −2 ft/s²). This means the acceleration continuously works against the motion, slowing the particle down. In fact, the particle will momentarily stop and then reverse direction — we will see this in the result.
Step 3
Substitute Values and Solve
Substituting s₀ = 0, v₀ = +12 ft/s, a = −2 ft/s², and t = 10 s:
\[
s = (0) + (12 \ \text{ft/s})(10 \ \text{s}) + \frac{1}{2}(-2 \ \text{ft/s}^2)(10 \ \text{s})^2
\]
\[
s = 0 + 120 – \frac{1}{2}(2)(100)
\]
\[
s = 120 – 100
\]
\[ s = 20 \ \text{ft} \]
Physical interpretation: The positive result confirms the particle is 20 ft to the right of the origin at t = 10 s. To understand the full motion, note that the particle first decelerates to a stop at t = v₀/|a| = 12/2 = 6 s (reaching s = 36 ft), then reverses and moves left, ending at s = 20 ft at t = 10 s. The particle traveled 36 ft outward and then 16 ft back — a total distance of 52 ft, but a net displacement of only +20 ft.
Final Answer
POSITION AT t = 10 s
s = 20 ft (to the right of the origin)