Rectilinear Kinematics
Velocity & Position from a Variable Acceleration Function
A particle begins its motion from rest along a straight-line path. Its acceleration varies with time according to
\( a = (2t – 6) \;\text{m/s}^2 \), where \( t \) is measured in seconds.
Determine the velocity of the particle at the instant \( t = 6 \;\text{s} \),
and find its position at \( t = 11 \;\text{s} \).
What We Know
- Initial velocity: \( v_0 = 0 \;\text{m/s} \) (starts from rest)
- Initial position: \( s_0 = 0 \;\text{m} \) (origin taken as reference)
- Acceleration function: \( a(t) = (2t – 6) \;\text{m/s}^2 \)
- Find velocity at \( t = 6 \;\text{s} \)
- Find position at \( t = 11 \;\text{s} \)
Why integration? — When acceleration is expressed as a function of time,
we cannot apply the constant-acceleration kinematic equations. Instead, we use the
fundamental definitions \( a = dv/dt \) and \( v = ds/dt \), then integrate each with
respect to time. This is the most direct and general approach for any \( a(t) \) problem.
By definition, acceleration is the time rate of change of velocity:
\[ a = \frac{dv}{dt} \quad \Longrightarrow \quad dv = a\, dt \]
Substituting the given expression \( a = (2t – 6) \) into the differential form gives us
a separable equation that connects infinitesimal changes in velocity to infinitesimal
changes in time:
\[ dv = (2t – 6)\, dt \]
Physical meaning: When \( t < 3\;\text{s} \), the acceleration is negative,
meaning the particle is being decelerated (even though it started from rest, it actually
moves in the negative direction initially). At \( t = 3\;\text{s} \), acceleration is zero —
this is an inflection point in the velocity curve.
We apply definite integration on both sides. The left side runs from the initial velocity
\( v_0 = 0 \) to the velocity \( v \) at some arbitrary time \( t \).
The right side runs from \( 0 \) to \( t \):
\[
\int_0^{v} dv = \int_0^{t} (2t – 6)\, dt
\]
Evaluating the right-hand integral term by term:
\[
\Big[v\Big]_0^{v} = \Big[t^2 – 6t\Big]_0^{t}
\]
\[
v – 0 = t^2 – 6t – 0
\]
\[ v(t) = t^2 – 6t \quad \text{(m/s)} \tag{1} \]
Check the shape: This is a parabola in time. Velocity is zero at \( t = 0 \)
(given) and again at \( t = 6\;\text{s} \). It reaches a minimum of \( -9\;\text{m/s} \)
at \( t = 3\;\text{s} \) — the particle moves backward before returning.
Substitute \( t = 6\;\text{s} \) into equation (1):
\[
v\big|_{t=6} = (6)^2 – 6(6) = 36 – 36
\]
\[ \boxed{v = 0 \;\text{m/s}} \]
Physical interpretation: At \( t = 6\;\text{s} \), the particle momentarily
comes to rest. This makes sense because the acceleration was negative from \( t = 0 \) to
\( t = 3\;\text{s} \) (particle slowing and reversing), then positive from \( t = 3\;\text{s} \)
onward (particle accelerating back in the positive direction). By \( t = 6\;\text{s} \),
it has exactly returned to rest before continuing to speed up.
Velocity is the time rate of change of position:
\[ v = \frac{ds}{dt} \quad \Longrightarrow \quad ds = v\, dt \]
Substituting the velocity function found in equation (1):
\[ ds = (t^2 – 6t)\, dt \]
This gives us another separable differential equation, this time linking position
to time. We integrate both sides with limits from \( s = 0 \) to \( s \)
and \( t = 0 \) to \( t \):
\[
\int_0^{s} ds = \int_0^{t} (t^2 – 6t)\, dt
\]
Evaluating each term:
\[
\Big[s\Big]_0^{s} = \left[\frac{t^3}{3} – 3t^2\right]_0^{t}
\]
\[
s – 0 = \frac{t^3}{3} – 3t^2 – 0
\]
\[ s(t) = \frac{t^3}{3} – 3t^2 \quad \text{(m)} \tag{2} \]
Note on direction: Because the velocity was negative between \( t = 0 \)
and \( t = 6\;\text{s} \), the particle traveled in the negative direction during
that interval. The position \( s \) here is the algebraic (signed) position, not
total distance traveled.
Substitute \( t = 11\;\text{s} \) into equation (2):
\[
s\big|_{t=11} = \frac{(11)^3}{3} – 3(11)^2
= \frac{1331}{3} – 3(121)
= 443.67 – 363
\]
\[ \boxed{s = 80.67 \;\text{m}} \]
What this means: At \( t = 11\;\text{s} \), the particle is located
80.67 m ahead of its starting point (in the positive direction). Even though it initially
moved backward, by \( t = 11\;\text{s} \) the forward motion has more than compensated,
and the net displacement is positive and substantial.
✦ Final Answers
Velocity at t = 6 s
\( v = 0 \;\text{m/s} \)
Position at t = 11 s
\( s = 80.67 \;\text{m} \)
