Maths with David

    9709. M1. General 1D Motion

    So far we have looked at conditions of constant acceleration. Of course, this is a modelling assumption and in reality moving particles typically experience change in acceleration (although falling particles under gravity are subject to constant acceleration if we ignore air resistance.

    As with velocity, we can think of acceleration as a function of time, which helps us to relate the three vector quantities: displacement, velocity and acceleration.

    As velocity is the change in displacement over time, we can think of instantaneous velocity in terms of differential calculus, such that v = \frac{ds}{dt} (using Newton’s original notation, v = \dot x and using Lagrange’s notation, v = s'(x)).

    Worked Examples

    • A particle moves in a straight line so that its displacement s at time t is given by s = t3 – 14t. Find an expression for its velocity at time t;
    • A ball moves in a straight line so that its displacement s at time t is given by s = 2t3 – 10t2. Find its speed when t=2;
    • A particle moves forwards and backwards along a straight line so that its displacement, s metres from the initial position, at time t seconds is given by s=2t3-12t2+18t. Find the distance that it travels in the first 5 seconds.

    Exercise

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    Acceleration, velocity and displacement – Differentiation

    Because acceleration is change in velocity over time and velocity is change in displacement over time, we can write them in the following way: a = \frac{dv}{dt} = \frac{d^2s}{dt^2} . Hence we can used differentiation to solve problems.

    Worked Examples

    • A particle moves in a straight line so that its velocity, v ms-1, at time t s is given by v=t2-4t. Find an expression for its acceleration at time t;
    • A motorcycle moves in a straight line so that its velocity, v ms-1, at time t s is given by v= 5t2-t3, for 0 < t < 4. Find its acceleration when t = 2;
    • A particle moves in a straight line so that its displacement, s m, at time t s (0≤t≤10) is given by s = \frac{1}{3}t^3 - 6t^2 + 15t
      • Sketch the shape of the velocity-time graph for the particle;
      • Hence, find the maximum speed of the particle.

    Exercise

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    Worked solutions

    Velocity and displacement – Integration

    Of course, as v = s'(t), so s = \int v(t) dt . Note that this is geometrically equivalent to the area under a curve velocity time graph giving displacement. Also note that in general s(0)=0, which can help us to find the constant of integration.

    Worked Examples

    • A particle moves in a straight line so that its velocity, v ms-1, at time t s is given by v = t3 – 4t. Find an expression for its displacement from the initial position at time t.
    • A particle moves in a straight line so that its velocity, v ms-1, at time t seconds after it starts to move is given by v = u + at, where u and a are constants. Find the displacement after t seconds.

    Exercise

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    Worked solutions

    Acceleration and Velocity – Integration

    The same relationship as between velocity and displacement applied between acceleration and velocity. As a = v'(t), so v = \int a(t) dt . Note that although generally s(0)=0, this does not apply with velocity, i.e. v(0) may not equal zero as an object may not initially be at rest. We need to know the velocity at some time in order to find the constant of integration.

    Worked Examples

    • A particle moves in a straight line so that its acceleration, a ms-2, at time t s is given by a = 8 – 4t. It starts with velocity 2 ms-1. Find an expression for its velocity at time t.
    • A particle moves in a straight line so that its acceleration, a ms-2, at time t s is given by a = 12 – 45t – 0.6t2. The particle comes to rest after 5 seconds.
      • Find the initial velocity of the particle;
      • Find the displacement after 2 seconds.

    Exercise & General Exercise

    Answers

    Worked solutions (2 exercises)