← Kinematic equations

Trips with more than one stage

The kinematic equations hold only while what is true of the motion?

Here is a car doing two things in one trip: it pulls away at 3 m/s and accelerates at 2 m/s² for 6 s — then the driver brakes at −5 m/s² until the car is moving at 5 m/s.

How much road does the braking take? No single equation covers this trip: you've already seen that each kinematic equation holds only while the acceleration is constant, and this trip's acceleration changes.

Split the motion into stages at every moment the acceleration changes. Within each stage the acceleration is constant, so the equations work — stage by stage.

stage 1: a = 2 m/s² for 6 sstage 2: a = −5 m/s² until 5 m/s3 m/s → ? m/s? m/s → 5 m/sthe velocity leaving stage 1 is the velocity entering stage 2
The bridge between stages: the final velocity of one stage is the starting velocity of the next.

Each stage is solved with the same steps as any other problem — values, equation, substitute, simplify. Stage 1 first: it hides the bridge velocity.

Write down the values:v = ? v₀ = 3 m/s a = 2 m/s² t = 6 sWrite down the equation:v = v₀ + atSubstitute in the values, and simplify:v = 3 + (2)(6) → v = 15 m/s

Stage 2 starts at that 15 m/s. Its own table has no time in it, so its own missing quantity picks its equation.

Write down the values:Δx = ? v = 5 m/s v₀ = 15 m/s a = −5 m/s²Write down the equation (no t anywhere):v² = v₀² + 2aΔxMake Δx the subject, substitute, simplify:Δx = (5² − 15²) / (2)(−5)Δx = (25 − 225) / (−10)Δx = 20 m

The braking takes 20 m of road — and every number in that answer came from running the ordinary routine twice.

A train moving at 8 m/s accelerates at 1.5 m/s² for 8 s, then slows steadily to 8 m/s over 56 m. What is its acceleration while slowing?

Find stage 1's final velocity (no Δx given or asked):

v = v₀ + at

v = 8 + (1.5)(8)

v = 20 m/s, carried into stage 2 as its starting velocity

Find stage 2's acceleration (no t given or asked). Make a the subject:

v² = v₀² + 2aΔx

a = (v² − v₀²) / 2Δx

a = (8² − 20²) / (2 × 56)

a = (64 − 400) / 112

a = −3 m/s² while slowing

A motorcycle moving at 2 m/s accelerates at 3 m/s² for 4 s, then brakes at −5 m/s² until it is moving at 4 m/s. How far does it travel while braking, in m? Enter the number only.

A speedboat moving at 5 m/s accelerates at 2.5 m/s² for 6 s, then slows at −4 m/s² until it is moving at 8 m/s. How long does the slowing stage take, in s? Enter the number only.

A bus moving at 10 m/s slows at −1.5 m/s² for 4 s in traffic, then accelerates at 3 m/s² for 4 s. What is its velocity at the end, in m/s? Enter the number only.

Back to the car at the start: it pulled away at 3 m/s, accelerated at 2 m/s² for 6 s to reach 15 m/s, then braked at −5 m/s² down to 5 m/s. We now know the braking took 20 m — one trip, two stages, and the same steps run twice.

Your turn

A car starts from rest, accelerates at 3 m/s² for 4 s, then brakes to rest over 18 m. What is its acceleration while braking, in m/s²? Include the sign in your answer.

A car moving at 6 m/s accelerates at 2 m/s² for 5 s, then slows steadily from that velocity to 4 m/s during 4 s. How far does it travel over the two stages combined, in m? Enter the number only.

A cyclist speeds up from 5 m/s to 13 m/s at 2 m/s², then slows from 13 m/s to 1 m/s at −3 m/s². How long do the two stages take in total, in s? Enter the number only.