Acceleration as a Function of t
When acceleration is given directly as a function of time — a thrust profile, a controlled deceleration, a time-varying engine — the problem reduces to two clean integrations. Integrate $a(t)$ to get $v(t)$, integrate $v(t)$ to get $x(t)$, and use the initial conditions at each step to pin down the constants. Aligned to NESA outcome MEX-M1.
If $a(t) = 4t$, $v(0) = 2$, $x(0) = 1$, what are $v(t)$ and $x(t)$? Before checking — integrate twice and apply each initial condition in turn.
When acceleration is a function of time alone, the chain $a \to v \to x$ becomes a chain of antiderivatives. Integrate $a(t)$ to get $v(t) + C_1$, apply the velocity initial condition, then integrate $v(t)$ to get $x(t) + C_2$ and apply the position initial condition. Two constants — two initial conditions.
The integrate-twice rule: $v(t) = \displaystyle\int a(t)\,dt + C_1$; $x(t) = \displaystyle\int v(t)\,dt + C_2$. Find $C_1$ from $v(0)$, then $C_2$ from $x(0)$ — in that order.
$a(t) \xrightarrow{\int dt} v(t) \xrightarrow{\int dt} x(t)$
Key facts
- $v(t) = \displaystyle\int a(t)\,dt + C_1$
- $x(t) = \displaystyle\int v(t)\,dt + C_2$
- Each integration introduces one arbitrary constant
- NESA outcome MEX-M1 covers this content
Concepts
- Why initial conditions must be applied in order (velocity, then position)
- Why "instantaneously at rest" means $v = 0$, not $a = 0$
- How a sign change in $v$ means the particle reverses direction
Skills
- Integrate any polynomial, trig or exponential $a(t)$
- Determine constants of integration from initial conditions
- Find when the particle is momentarily at rest, and the corresponding position
Given $a(t)$ and two initial conditions, the procedure is:
- Integrate $a(t)$ with respect to $t$; call the result $v(t) + C_1$.
- Apply the velocity initial condition (e.g. $v(0) = v_0$) to evaluate $C_1$.
- Integrate the now-known $v(t)$ to get $x(t) + C_2$.
- Apply the position initial condition (e.g. $x(0) = x_0$) to evaluate $C_2$.
Integrate $a(t)$ once: get $v$ (with $C_1$). Use $v(0)$ to find $C_1$. · Integrate $v(t)$ once: get $x$ (with $C_2$). Use $x(0)$ to find $C_2$. · Always apply ICs in order (v before x) and label $C_1$, $C_2$ clearly. · Definite-integral form $v(t) = v_0 + \int_0^t a(s)\,ds$ skips constants entirely.
Pause — copy the integrate-twice procedure, the IC-application order ($v(0)$ fixes $C_1$, $x(0)$ fixes $C_2$), and the definite-integral form $v(t) = v_0 + \int_0^t a(s)\,ds$ into your book.
Quick check: A particle has $a(t) = 6t$, $v(0) = 1$ and $x(0) = 0$. What is $x(t)$?
We just saw the integrate-twice procedure: $v(t) = v_0 + \int_0^t a(s)\,ds$ and $x(t) = x_0 + \int_0^t v(s)\,ds$, with each IC applied immediately after its integration. That raises a question: how do we extract information about when the particle is at rest, reverses direction, or covers a total distance? This card answers it → at rest means $v(t) = 0$; sign change in $v$ means reversal; total distance requires $\int|v|\,dt$, split at rest times.
Two diagnostic questions about a time-varying motion:
- When is the particle at rest? Set $v(t) = 0$ and solve for $t$. ($a = 0$ is not the same — that's an inflection in $v$, not a rest.)
- Does the particle reverse? If $v$ changes sign at $t = t_0$, the particle is momentarily at rest at $t_0$ and then moves the other way.
Distance vs displacement. Displacement from $t_1$ to $t_2$ is $x(t_2) - x(t_1)$. Total distance is the sum of absolute displacements between consecutive rest points:
"At rest" means $v(t) = 0$, NOT $a(t) = 0$ · Sign change in $v$ ⇒ reversal of motion direction · Distance = $\int|v|\,dt$ — split at rest times if $v$ changes sign · Displacement = $x(t_2) - x(t_1)$, signed
Pause — copy the at-rest condition $v(t) = 0$ (not $a(t) = 0$), the reversal criterion (sign change in $v$), $d = \int|v|\,dt$ for total distance, and displacement = $x(t_2)-x(t_1)$ into your book.
Did you get this? True or false: a particle is momentarily at rest exactly when its acceleration is zero.
Worked examples · 3 in a row, reveal as you go
A particle has $a(t) = 12t - 6$ (m/s$^2$) with $v(0) = 0$ and $x(0) = 4$. Find $v(t)$ and $x(t)$.
A rocket has $a(t) = 6 - 0.5t$ (m/s$^2$) with $v(0) = 0$ and $x(0) = 0$. Find the time of maximum velocity, the maximum velocity, and the position at that time.
A particle has $a(t) = -4\cos 2t$ (m/s$^2$) with $v(0) = 0$ and $x(0) = 1$. Find $v(t)$, $x(t)$, the times of momentary rest, and the total distance travelled between $t = 0$ and $t = \pi$.
Fill the gap: Given $a(t)$, the velocity is $v(t) = \displaystyle\int a(t)\,dt + $, then the position is $x(t) = \displaystyle\int v(t)\,dt + $, with each constant fixed by an initial condition.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for any $a(t)$, you need exactly two initial conditions to determine both $v(t)$ and $x(t)$ uniquely.
Activities · practice with the ideas
$a(t) = 6t - 2$, $v(0) = 1$, $x(0) = 0$. Find $v(t)$ and $x(t)$.
$a(t) = e^{-t}$, $v(0) = 2$, $x(0) = 1$. Find $v(t)$ and $x(t)$. What is the limiting velocity as $t \to \infty$?
$a(t) = \sin t$, $v(0) = 1$, $x(0) = 0$. Show the motion is bounded and find the total distance from $t = 0$ to $t = 2\pi$.
$a(t) = 2 - t$ with $v(0) = 0$, $x(0) = 0$. Find the time when the particle is momentarily at rest and its position then.
$a(t) = -4\sin 2t$ with $v(0) = 2$ and $x(0) = 0$. Find $v(t)$, $x(t)$, and verify that the motion satisfies $\ddot x = -4x$.
Odd one out: Three of these correctly relate $a$, $v$, $x$ in time. Which one is NOT a valid identity?
At the start you predicted what happens to a rocket with $a(t) = 6 - 0.5t$, $v(0) = 0$, $x(0) = 0$ — when acceleration vanishes, the maximum velocity, whether it eventually stops.
From worked example 2: $v(t) = 6t - 0.25t^2$. Acceleration is zero at $t = 12$ s, where velocity peaks at $36$ m/s and position is $288$ m. After that, $a < 0$ so velocity decreases, eventually reaching zero again at $t = 24$ s — and then becoming negative (the rocket reverses). The key insight: peak speed is not when the rocket reaches max altitude, and "still accelerating" is not the same as "still speeding up" once the sign of $a$ flips.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A particle has $a(t) = 4t + 2$ with $v(0) = 3$ and $x(0) = 0$. Find $v(t)$ and $x(t)$. (2 marks)
Q2. A particle has $a(t) = 3 - t$ (m/s$^2$), $v(0) = 0$, $x(0) = 0$. (a) Find the time at which the particle is momentarily at rest after $t = 0$. (b) Find the position then. (3 marks)
Q3. A particle has $a(t) = -9\sin 3t$, $v(0) = 3$, $x(0) = 0$. (a) Find $v(t)$ and $x(t)$. (b) Show that the motion satisfies $\ddot x + 9x = 0$ (i.e. SHM with $\omega = 3$). (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $v(t) = 3t^2 - 2t + C_1$; $v(0) = 1 \Rightarrow C_1 = 1$, so $v(t) = 3t^2 - 2t + 1$. $x(t) = t^3 - t^2 + t + C_2$; $x(0) = 0 \Rightarrow C_2 = 0$, so $x(t) = t^3 - t^2 + t$.
2. $v(t) = -e^{-t} + C_1$; $v(0) = 2 \Rightarrow -1 + C_1 = 2 \Rightarrow C_1 = 3$, so $v(t) = 3 - e^{-t}$. Limit: $v \to 3$ as $t \to \infty$. $x(t) = 3t + e^{-t} + C_2$; $x(0) = 0 + 1 + C_2 = 1 \Rightarrow C_2 = 0$, so $x(t) = 3t + e^{-t}$.
3. $v(t) = -\cos t + C_1$; $v(0) = 1 \Rightarrow C_1 = 2$, so $v(t) = 2 - \cos t$, always between 1 and 3 (positive). No rest on $[0, 2\pi]$. Distance $= \displaystyle\int_0^{2\pi} (2 - \cos t)\,dt = [2t - \sin t]_0^{2\pi} = 4\pi$ m.
4. $v(t) = 2t - t^2/2$. Set $v = 0$: $t(2 - t/2) = 0 \Rightarrow t = 0$ or $t = 4$. Rest at $t = 4$. $x(t) = t^2 - t^3/6$; $x(4) = 16 - 64/6 = 32/3$ m.
5. $v(t) = 2\cos 2t + C_1$; $v(0) = 2 \Rightarrow C_1 = 0 \Rightarrow v = 2\cos 2t$. $x(t) = \sin 2t + C_2$; $x(0) = 0 \Rightarrow C_2 = 0$, so $x = \sin 2t$. Then $\ddot x = -4\sin 2t = -4x$, confirming SHM.
Q1 (2 marks): $v(t) = 2t^2 + 2t + 3$ [1]; $x(t) = \tfrac{2}{3}t^3 + t^2 + 3t$ [1].
Q2 (3 marks): $v(t) = 3t - t^2/2$ [1]. Rest: $v = 0 \Rightarrow t = 6$ s (after $t=0$) [1]. $x(t) = \tfrac{3}{2}t^2 - t^3/6$; $x(6) = 54 - 36 = 18$ m [1].
Q3 (3 marks): $v(t) = 3\cos 3t$, $x(t) = \sin 3t$ [1]. Compute $\ddot x = -9\sin 3t$ [1]. Hence $\ddot x + 9x = -9\sin 3t + 9\sin 3t = 0$, SHM with $\omega = 3$ [1].
Five timed questions on $a(t) \to v(t) \to x(t)$ with initial conditions, rest times, and reversal. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick integrate-twice questions. Lighter alternative to the boss.
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