Difficulty: Intermediate | Prerequisites: Antiderivatives and the power rule (§3.9), basic kinematics vocabulary.
This topic applies antiderivatives to physics: given acceleration, recover velocity; given velocity, recover position. The extra ingredient is an initial condition (a known value at a specific time), which pins down the constant C. These "initial value problems" are the first place in the course where calculus solves a concrete physical problem. If you can differentiate, you already know the forward direction; this is the reverse.
To find velocity from acceleration, antidifferentiate and use v(0) to solve for C. To find position from velocity, antidifferentiate again and use s(0) to solve for the new C. Stopping-distance and projectile problems are standard applications of this two-step process.
Acceleration, a(t)
The rate of change of velocity with respect to time. In calculus terms, a(t) = v'(t) = s''(t).
Velocity, v(t)
The rate of change of position with respect to time. v(t) = s'(t). It is the antiderivative of acceleration.
Position (displacement), s(t)
The location of an object along a coordinate line at time t. It is the antiderivative of velocity.
Initial condition
A known value of the function at a specific time, such as v(0) = 4 or s(0) = 2. This fixes the constant of integration C.
Initial value problem (IVP)
A problem that asks you to find a specific antiderivative satisfying a given initial condition, rather than the general antiderivative. Think of it as: the initial condition picks out one curve from the whole family F(x) + C.
Free fall / gravitational acceleration
Motion under gravity alone, where a(t) = −9.8 m/s² (metric) or a(t) = −32 ft/s² (imperial). The negative sign means gravity pulls downward.
Deceleration
Negative acceleration, meaning the object is slowing down. In braking problems, the acceleration is opposite to the direction of motion.
Start with a(t). Antidifferentiate to get v(t) + C₁.
Use the initial velocity v(0) = v₀ to solve for C₁.
Antidifferentiate v(t) to get s(t) + C₂.
Use the initial position s(0) = s₀ to solve for C₂.
This chain (acceleration → velocity → position) is the backbone of every motion problem in this section.
Given a(t) = (t + 2)³ ft/s², v(0) = 4 ft/s, s(0) = 2 ft.
Finding v(t):
Antidifferentiate: v(t) = (t + 2)⁴ / 4 + C
Apply v(0) = 4: (0 + 2)⁴ / 4 + C = 4, so 16/4 + C = 4, giving C = 0
Result: v(t) = (t + 2)⁴ / 4
Finding s(t):
Antidifferentiate: s(t) = (t + 2)⁵ / 20 + C
Apply s(0) = 2: (0 + 2)⁵ / 20 + C = 2, so 32/20 + C = 2, giving C = 2/5
Result: s(t) = (t + 2)⁵ / 20 + 2/5
A car travels at 60 km/h and decelerates at 10 m/s².
Unit conversion first: 60 km/h = 60 × 1000/3600 = 16.667 m/s.
Finding the stopping time:
v(t) = v₀ + a · t = 16.667 + (−10)t
Set v = 0: t = 16.667 / 10 = 1.667 seconds
Finding the stopping distance:
Δx = (v₀ + 0)/2 × t = (0 + 16.667)/2 × 1.667 ≈ 13.89 m
Alternatively, antidifferentiate v(t) to get s(t) and evaluate.
Two runners accelerate at different rates up to different top speeds.
Runner 1 (you): a = 3 m/s², top speed = 10 m/s
Acceleration phase: v = 3t until v = 10, so t = 10/3 s
Distance during acceleration: s = (3/2)t² = (3/2)(10/3)² = 50/3 m
Remaining distance at top speed: 100 − 50/3 = 250/3 m at 10 m/s → takes 25/3 s
Total time: 10/3 + 25/3 = 35/3 ≈ 11.67 s
Runner 2 (friend): a = 2 m/s², top speed = 12 m/s
Acceleration phase: v = 2t until v = 12, so t = 6 s
Distance during acceleration: s = t² = 36 m
Remaining distance at top speed: 100 − 36 = 64 m at 12 m/s → takes 64/12 = 16/3 s
Total time: 6 + 16/3 = 34/3 ≈ 11.33 s
The friend finishes first, despite slower acceleration, because of the higher top speed.
A ball is dropped from 200 m. Position: s(t) = 200 − 4.9t².
Time to hit ground: Set s(t) = 0 → t² = 200/4.9 → t ≈ 6.39 s.
Speed at impact: v(t) = s'(t) = −9.8t → v(6.39) ≈ −62.6 m/s (the negative sign indicates downward motion; speed is 62.6 m/s).
With a(t) = −9.8 m/s² and s(0) = 0 (ground level):
v(t) = −9.8t + v₀
s(t) = −4.9t² + v₀t
Maximum height occurs when v(t) = 0, i.e. t = v₀/9.8
Maximum s = −4.9(v₀/9.8)² + v₀(v₀/9.8) = v₀²/19.6
Set v₀²/19.6 = 200 → v₀ = √(3920) ≈ 62.6 m/s
You must throw the ball upward at approximately 62.6 m/s to reach 200 m.
v(t) = \int a(t) \, dts(t) = \int v(t) \, dt\text{Free fall (metric): } a(t) = -9.8 \text{ m/s}^2\text{Free fall (imperial): } a(t) = -32 \text{ ft/s}^2\text{Max height for vertical throw: } s_{\max} = \frac{v_0^2}{2 \times 9.8}\text{Constant accel stopping distance: } \Delta x = \frac{v_0^2}{2|a|}Stopping-distance calculations are used by traffic engineers to set speed limits and design safe following distances. Projectile motion governs everything from sports (how fast to throw a ball to clear a wall) to aerospace (launch velocities for rockets). Accelerometers in your phone integrate acceleration data to estimate velocity and displacement, using exactly this chain of antiderivatives.
Students often forget to convert units before computing. Mixing km/h with m/s² produces nonsense. Always convert to consistent units first.
Forgetting the initial condition step is a classic error. The general antiderivative has a C; without plugging in the initial condition, your answer is incomplete.
Students confuse "speed" and "velocity." Speed is |v(t)|, always non-negative. Velocity can be negative (indicating direction). When a problem asks "how fast," it usually wants speed.
In projectile problems, students often drop the negative sign on gravitational acceleration. If upward is positive, then a = −9.8 m/s², not +9.8.
⚠️ The acceleration-to-velocity-to-position chain with initial conditions is a near-certain exam topic. Practise the full two-step process until it is automatic.
⚠️ Unit conversion errors cost easy marks. If the problem mixes km/h and m/s², convert before you begin.
⚠️ Stopping-distance and braking problems are popular word problems. Know the formula Δx = v₀²/(2|a|) as a shortcut, but be ready to derive it from antidifferentiation.
⚠️ Free-fall problems test whether you handle the sign of acceleration correctly and whether you can set up the equation for maximum height (v = 0 at the peak).
True or false: If a(t) = 5 and v(0) = 3, then v(t) = 5t + 3. (True.)
Fill in the blank: The position function is the antiderivative of the ______ function. (velocity)
True or false: A ball thrown upward at 20 m/s reaches maximum height when v(t) = 20. (False, it reaches max height when v(t) = 0.)
Fill in the blank: 60 km/h = ______ m/s. (16.667 m/s)
True or false: In free fall near Earth's surface, a(t) = +9.8 m/s² if we take upward as positive. (False, a(t) = −9.8 m/s².)
Q: An object has acceleration a(t) = 6t and initial velocity v(0) = 2. Find v(t).
A: v(t) = ∫ 6t dt = 3t² + C. Using v(0) = 2: C = 2. So v(t) = 3t² + 2.
Q: Continuing the above, if s(0) = 0, find s(t).
A: s(t) = ∫ (3t² + 2) dt = t³ + 2t + C. Using s(0) = 0: C = 0. So s(t) = t³ + 2t.
Q: A car brakes from 25 m/s with deceleration 5 m/s². How far does it travel before stopping?
A: Stopping time: t = 25/5 = 5 s. Distance: Δx = v₀²/(2a) = 625/10 = 62.5 m.
Q: A ball is thrown upward from ground level with initial velocity v₀. Derive a formula for the maximum height in terms of v₀.
A: v(t) = −9.8t + v₀. At max height v = 0, so t = v₀/9.8. Then s(t) = −4.9(v₀/9.8)² + v₀(v₀/9.8) = v₀²/(2 × 9.8) = v₀²/19.6.
Q: An object moves with a(t) = (t + 2)³ ft/s², v(0) = 4 ft/s. Find v(t).
A: v(t) = (t + 2)⁴/4 + C. v(0) = 16/4 + C = 4, so C = 0. Therefore v(t) = (t + 2)⁴/4.
This topic builds directly on antiderivatives (§3.9 Part 1). The same initial-value framework returns in differential equations courses. The relationship a(t) → v(t) → s(t) is a special case of the Fundamental Theorem of Calculus (Chapter 5), which generalises the idea that the integral of a rate gives the net change. Piecewise motion problems (like the race example) preview piecewise-defined integrals.
Rectilinear motion, initial value problem, IVP, velocity from acceleration, position from velocity, stopping distance, braking distance, free fall, projectile motion, gravitational acceleration, constant of integration, kinematics, MTH 132, Calculus I, Section 3.9