A Complete Guide to Kinematics with Constant Acceleration
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Acceleration describes how quickly velocity changes.
Uniformly accelerated motion means acceleration a stays constant throughout the motion. Velocity changes by equal amounts in equal time intervals.
Units: m/s² — every second, velocity increases (or decreases) by a m/s.
An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a net external force.
If velocity is changing, there must be a net force. Uniform acceleration ⇒ constant net force.
If F and m are both constant, then a is constant — giving us uniformly accelerated motion, and all the kinematic formulas that follow.
When acceleration a is constant:
| Variable | Meaning |
|---|---|
| v₀ | Initial velocity |
| v | Final velocity |
| a | Constant acceleration |
| t | Time elapsed |
| s | Displacement |
Each formula omits one variable — pick the formula that contains exactly what you know + what you want.
Slope = acceleration | Area under graph = displacement
Drag the sliders to explore!
Given initial conditions + acceleration/force → find final state.
A car accelerates uniformly from rest. After time t, its speed is v. How long does it take for its kinetic energy to become 9× the value at time t?
Solution: KE ∝ v². For KE to be 9×, speed must be 3v. Starting from rest with constant a: v = at → t₁ = v/a. For 3v: t₂ = 3v/a = 3t₁. Answer: 3t
A train of length L accelerates uniformly from rest. Time for the whole train to pass a pole is T. Find the time for the last carriage (length l) to pass.
Solution: s = ½at². Whole train: L = ½aT². Front (L−l) passes: L−l = ½at₁². Last carriage time = T − t₁ = T − T√((L−l)/L) = T(1 − √(1 − l/L))
For an object starting from rest (v₀ = 0) with constant acceleration:
s(total after nT) = ½a(nT)² = n² · ½aT². So sₙ = s(nT) − s((n−1)T) = (2n−1) · ½aT². Ratio = 1 : 3 : 5 : 7 …
Velocity cannot go negative in braking! Once v = 0, the object stops. Do NOT blindly plug large t into formulas — check if the object has already stopped.
A car travels at 20 m/s and brakes with a = −5 m/s². Find displacement in the first 5 seconds.
Step 1: Time to stop: 0 = 20 + (−5)t → t_stop = 4 s < 5 s!
Step 2: The car stops at t = 4 s. s = 20(4) + ½(−5)(4²) = 80 − 40 = 40 m
Common mistake: Using t = 5 gives s = 20(5) + ½(−5)(25) = 37.5 m (wrong!)
Sometimes it's easier to think backwards from the end of the motion.
An object decelerating to rest is the reverse of one accelerating from rest. Use this symmetry!
A car brakes uniformly and covers 15 m in the last second before stopping. What was its speed 1 s before stopping?
Reverse: Think of it as starting from rest and accelerating. In the 1st second from rest: s = ½a(1)² = 15 → a = 30 m/s². Speed at 1 s: v = 30(1) = 30 m/s
A car brakes and stops after 50 m. The distance in the last 1 s is 1 m. Find total braking time.
Reverse: ½a(1)² = 1 → a = 2 m/s². Total: 50 = ½(2)t² → t = √50 ≈ 7.07 s
Ratio of distances in the last three equal intervals before stopping = 5 : 3 : 1 (reverse of 1 : 3 : 5).
Objects often accelerate, cruise, then decelerate — a trapezoidal v-t graph.
Adjust sliders to see total displacement.
When a problem uses letters instead of numbers, the approach is the same — but answers are algebraic expressions.
A particle starts from rest and accelerates at a₁ for time T, then decelerates at a₂ until it stops. Find the total distance.
Phase 1: v_max = a₁T, s₁ = ½a₁T²
Phase 2: Time to stop: t₂ = a₁T/a₂, s₂ = v_max²/(2a₂) = a₁²T²/(2a₂)
Total: s = ½a₁T² + a₁²T²/(2a₂) = ½a₁T²(1 + a₁/a₂)
| Tool | When to Use | Formula / Method |
|---|---|---|
| Basic kinematics | Know 3 of {v₀, v, a, t, s} | Pick the formula missing the unknown you don't need |
| Average velocity | Quick displacement calc | s = ½(v₀+v)·t |
| v-t graph | Visualize multi-stage; find displacement as area | Slope = a, Area = s |
| Odd-number ratio | Equal time intervals from rest | 1 : 3 : 5 : 7 … |
| Reverse thinking | Braking / deceleration to rest | Treat as acceleration from rest, reversed |
| Braking check | Any deceleration problem | Find t_stop first; clamp t ≤ t_stop |
| Phase splitting | Multi-stage motion | Link phases via shared v or s |
| Dimensional analysis | Parametric answers | Check units match on both sides |
A rocket accelerates from rest at 20 m/s² for 10 s, then the engine shuts off and it decelerates at 10 m/s² until it stops. Find: (a) maximum speed, (b) total distance, (c) total time, (d) average speed for the entire journey.
Core formulas + forward problems (Slides 2-7)
Odd-number ratio + braking (Slides 8-9)
Reverse thinking + multi-stage (Slides 10-11)
Parametric problems + mixed practice (Slide 12)
Timed test: 10 problems in 40 min, review mistakes