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A.1 · 8 · A.1 Kinematics · applying v = u + at, x = ut + ½at², v² = u² + 2as

Exam questions I: a train, a bouncing ball, a skidding car

Three exam-style motion problems, worked step by step with our own animated diagrams: read the graph, then check it against the moving object itself.

SL+HLHarder28 minFree
Hook

A graph tells you the numbers. An animation tells you what those numbers actually look like happening. Here are three classic exam problems, solved both ways.

Problem 1 · speeding up, then slowing down
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A train from rest to rest

A train starts from rest at station A and speeds up steadily until it passes point B. From B it slows down steadily until it comes to rest at station C. The whole journey, A to C, takes time T; the part from A to B takes 0.80 of that time.

GivenValue
Acceleration, A → B0.20 m s⁻²
Fraction of the journey time spent A → B0.80 T
Total distance, A → C1800 m
Speed at A and at C0 (starts and ends at rest)
Simulation

Watch the train, and the speed readout

The idea: Scrub through the journey. The train speeds up for the first 80% of the time, then must lose ALL of that speed in only the remaining 20%.

Predict: since the deceleration phase (B → C) is much SHORTER than the acceleration phase (A → B), but must remove the same speed, the deceleration compared to the 0.20 m s⁻² acceleration is…

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

WORKED EXAMPLE · FIND T, THE SPEED AT B, AND THE DECELERATION

A train starts from rest at A, accelerates at 0.20 m s⁻² until it passes B (reached after 0.80 of the total journey time T), then decelerates uniformly to rest at C. The whole A-to-C journey covers 1800 m. On a v–t graph this is a triangle: base T, with the peak (at B) reached after 0.80T.

    Check 1 of 6

    On the train’s v–t graph, the AREA under the whole triangle represents…

    Check 2 of 6

    A train accelerates for a SHORTER time than it decelerates, but reaches the same peak speed both times. Its acceleration, compared with its deceleration, is…

    Check 3 of 6

    A ball bounces and loses some speed each time. Each successive rebound height is…

    Check 4 of 6

    At the very top of a ball’s bounce, its velocity is zero. Its acceleration at that instant is…

    Check 5 of 6

    In the skid-mark problem, the car’s speed during the REACTION phase (before braking) is…

    Check 6 of 6

    Why do skid marks only appear during the braking phase, not the reaction phase?

    Apply

    A different train accelerates from rest at 0.25 m s⁻² for the first 60% of its journey time T, then decelerates to rest for the remaining 40%. If the total distance is 2400 m, find T (set up the same way as Problem 1, then solve).

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    Remember earlier lessons
    From lesson A.1 · 7

    The area under a v–t graph is…

    From lesson A.1 · 6

    v = u + at rearranges v² = u² + 2as by…

    Summary card

    Key points

    • • A v–t graph shaped like a triangle: its area is the total displacement, its two slopes are the two accelerations.
    • • A shorter phase reaching the same speed change needs a BIGGER magnitude of acceleration.
    • • At the top of a bounce, velocity is zero but acceleration (g) is not — they are independent.
    • • A "reaction, then brake" problem is two separate uniform phases: constant velocity, then constant deceleration; solve the braking phase first with v² = u² + 2as, since it pins down u.

    Formulas

    • v = u + at
    • x = ut + ½at²
    • v² = u² + 2as
    • area under v–t = displacement

    Key terms

    Uniform deceleration
    :
    constant negative acceleration, decreasing speed at a constant rate

    Common errors

    • • Assuming both phases of a "speeds up then slows down" graph must have the same acceleration.
    • • Thinking zero velocity means zero acceleration.
    • • Applying the braking deceleration to the reaction-time distance, or vice versa.
    Checks solved: 0 of 6
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