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A.1 · 9 · A.1 Kinematics · applying v = u + at, x = ut + ½at², v² = u² + 2as

Exam questions II: a signal, a take-off, a chase, and leaking oil

Four more exam-style motion problems as multiple-choice, each with its own recreated diagram, animation and worked solution.

SL+HLHarder26 minFree
Hook

Four more scenes: a train that must not run a red signal, a plane with a limited runway, one car catching another, and a leaking car whose oil drops become a stopwatch.

Problem 4 · brake in time
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A train must stop by the red signal

A train driver applies the brakes at a yellow signal, a known distance from a red signal where the train must be at rest. Assuming the greatest deceleration the train can manage, what is the fastest it could safely be going at the yellow signal?

GivenValue
Distance, yellow signal to red signal1000 m
Maximum deceleration0.20 m s⁻²
Speed required at the red signal0 (must have stopped)
Simulation

The fastest safe approach speed

The idea: This train is travelling at exactly the fastest safe speed: it comes to rest exactly at the red signal, using the full 1000 m.

Predict: on a particular railway, a train driver applies the brake at a yellow signal, a distance of 1.0 km from a red signal where it stops. The maximum deceleration is 0.2 m s⁻². Assuming uniform deceleration, what is the maximum safe speed at the yellow signal?

Answer the prediction above to unlock the simulation. Then test it.

WORKED EXAMPLE · THE MAXIMUM SAFE SPEED

A train driver applies the brake at a yellow signal, 1000 m from a red signal where the train must have already stopped. The maximum deceleration is 0.20 m s⁻², applied uniformly. Use v² = u² + 2as, with the final speed v = 0, s = 1000 m and a = −0.20 m s⁻².

    Check 1 of 4

    A car brakes from speed u and comes exactly to rest over a known distance s, at a known deceleration a. Which equation finds u directly, with no need for time?

    Check 2 of 4

    Before using v² = u² + 2as with a speed given in km h⁻¹, you should…

    Check 3 of 4

    Car Y starts a distance d behind car X and eventually draws level. At that moment, the distance Y has travelled equals…

    Check 4 of 4

    For a uniformly accelerating object, the average velocity across a time interval equals the instantaneous velocity…

    Apply

    A cyclist must stop within 40 m of applying the brakes, decelerating uniformly at 2.5 m s⁻². What is the fastest safe speed at the moment the brakes are applied?

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    Remember earlier lessons
    From lesson A.1 · 8

    A v–t graph shaped like a triangle: its area is…

    From lesson A.1 · 6

    v² = u² + 2as is useful when you do not know…

    Summary card

    Key points

    • • "Just enough" problems (a signal, a runway) both come down to v² = u² + 2as with one speed equal to zero.
    • • Always convert to consistent SI units (m, s, m s⁻¹) before using a kinematics equation.
    • • In a chase problem, the gap equals the difference between the two distances travelled by the time they meet.
    • • For uniform acceleration, average velocity across an interval = instantaneous velocity at its midpoint — this is how ticker-tape and oil-drop spacing reveal acceleration.

    Formulas

    • v² = u² + 2as
    • v = u + at
    • x = ut + ½at²

    Key terms

    Midpoint velocity
    :
    for uniform acceleration, the average velocity over an interval, which equals the instantaneous velocity halfway through it in time

    Common errors

    • • Using a speed in km h⁻¹ directly in a kinematics equation.
    • • Forgetting the initial gap when two vehicles are said to "meet" or "draw level".
    • • Treating average velocity as if it applied at the START or END of an interval instead of its midpoint.
    Checks solved: 0 of 4
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