Rectilinear Kinematics
Starting from Rest with Variable Acceleration
A particle begins its motion from a state of complete rest and travels along a straight-line path.
Its acceleration varies with time according to the expression a = (2t − 6) m/s²,
where t is measured in seconds. Determine the velocity of the particle at t = 6 s,
and find its position along the path at t = 11 s.
What We Know
- Initial velocity: v₀ = 0 m/s (starts from rest)
- Initial position: s₀ = 0 m (origin at start point)
- Acceleration function: a(t) = (2t − 6) m/s²
- Required: velocity at t = 6 s
- Required: position at t = 11 s
Why integration? The acceleration here is a function of time, not a constant — so the standard constant-acceleration formulas do not apply. Instead we use the fundamental definitions a = dv/dt and v = ds/dt, separate the variables, and integrate step by step: acceleration → velocity → position.
Step 1
Set Up the Velocity Integral
By definition, acceleration is the time rate of change of velocity. Rearranging separates the variables v and t onto opposite sides, making integration straightforward:
\[ a = \frac{dv}{dt} \quad \Longrightarrow \quad dv = a \, dt \]
Substituting the given acceleration expression:
\[ dv = (2t – 6) \, dt \]
Physical note: The acceleration a = 2t − 6 is negative when t < 3 s, zero at t = 3 s, and positive when t > 3 s. The particle initially slows down (in the negative direction), momentarily stops accelerating at t = 3 s, then begins gaining speed in the positive direction.
Step 2
Integrate to Find v(t) and Evaluate at t = 6 s
Apply definite integration using the initial condition v = 0 at t = 0, leaving the upper limits as the general values v and t:
\[ \int_{0}^{v} dv = \int_{0}^{t} (2t – 6) \, dt \]
\[ \Big[v\Big]_{0}^{v} = \Big[t^2 – 6t\Big]_{0}^{t} \]
\[ v – 0 = t^2 – 6t – 0 \]
\[ v(t) = t^2 – 6t \quad \text{m/s} \qquad \cdots (1) \]
Substituting t = 6 s into equation (1):
\[ v\big|_{t=6} = (6)^2 – 6(6) = 36 – 36 \]
\[ v = 0 \ \text{m/s} \]
Physical interpretation: Velocity = 0 at t = 6 s does not mean the particle was stationary throughout. It started from rest, built up negative velocity, then the increasing positive acceleration reversed that motion — the particle passes back through zero velocity at t = 6 s before continuing in the positive direction.
Step 3
Set Up the Position Integral
Velocity is the time rate of change of position. Separating variables gives:
\[ v = \frac{ds}{dt} \quad \Longrightarrow \quad ds = v \, dt \]
Substituting the velocity expression from equation (1):
\[ ds = (t^2 – 6t) \, dt \]
Why integrate again? Each integration step climbs one level in the kinematic chain: acceleration → velocity → position. Because a(t) is a polynomial, each integration simply raises the polynomial degree by one — a clean, systematic process.
Step 4
Integrate to Find s(t) and Evaluate at t = 11 s
Apply definite integration using the initial condition s = 0 at t = 0:
\[ \int_{0}^{s} ds = \int_{0}^{t} (t^2 – 6t) \, dt \]
\[ \Big[s\Big]_{0}^{s} = \left[\frac{t^3}{3} – 3t^2\right]_{0}^{t} \]
\[ s = \frac{t^3}{3} – 3t^2 \]
\[ s(t) = \frac{t^3}{3} – 3t^2 \quad \text{m} \qquad \cdots (2) \]
Substituting t = 11 s into equation (2):
\[ s\big|_{t=11} = \frac{(11)^3}{3} – 3(11)^2 = \frac{1331}{3} – 363 = 443.67 – 363 \]
\[ s = 80.67 \ \text{m} \]
Physical interpretation: The positive value confirms the particle ends up 80.67 m ahead of its starting point. Although it moved in the negative direction during the early phase (0 < t < 6 s), the strong positive acceleration phase afterward more than compensates, giving a net positive displacement.
Final Answers
VELOCITY AT t = 6 s
v = 0 m/s
POSITION AT t = 11 s
s = 80.67 m