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Kinematics using calculusEdexcel International A Level Further Maths: Flashcards

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What is velocity in terms of displacement?

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What is velocity in terms of displacement?
v=dxdtv=\dfrac{dx}{dt}
What is acceleration in terms of velocity and displacement?
a=dvdt=d2xdt2a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}
How do you find velocity from acceleration a(t)a(t)?
Integrate: v=∫a dtv=\int a\,dt, then use an initial condition for the constant.
How do you find when a particle is at rest?
Solve v=0v=0.
How do you find the time of greatest velocity?
Solve a=0a=0, then check the end points.
Can you use suvat when aa depends on tt?
No; suvat needs constant acceleration.
What is the difference between speed and velocity?
Speed is the magnitude of velocity and is never negative.
How do you find the total distance when the particle changes direction?
Split at each time v=0v=0 and add the magnitudes of the separate displacements.
What does ∫t1t2v dt\int_{t_1}^{t_2}v\,dt give?
The displacement between t1t_1 and t2t_2.
Velocity from a position vector r\mathbf{r}?
v=drdt\mathbf{v}=\dfrac{d\mathbf{r}}{dt}, differentiating each component.
How do you find the speed from a velocity vector?
∣v∣=vx2+vy2|\mathbf{v}|=\sqrt{v_x^2+v_y^2}
r=t3i+2t2j\mathbf{r}=t^3\mathbf{i}+2t^2\mathbf{j}: find v\mathbf{v} and a\mathbf{a}.
v=3t2i+4tj\mathbf{v}=3t^2\mathbf{i}+4t\mathbf{j}, a=6ti+4j\mathbf{a}=6t\mathbf{i}+4\mathbf{j}
Why must you include a constant when integrating?
Initial conditions fix the constant; omitting it gives the wrong velocity or displacement.

Exam questions on Kinematics using calculus

  1. A particle PP moves along a straight line. Its displacement xx metres from a fixed point OO at time tt seconds is given by x=t3−6t2+9tx=t^3-6t^2+9t, for t≥0t\geq0.
    Find the displacement of PP from OO at the instant when its acceleration is zero.2 marks
  2. A particle QQ moves along a straight line through a fixed point OO. At time tt seconds its velocity is v=6t−3t2v=6t-3t^2 in m s−1\text{m s}^{-1}, for 0≤t≤30\leq t\leq3, and QQ is at OO when t=0t=0.
    Find the greatest velocity of QQ for 0≤t≤30\leq t\leq3.2 marks
  3. A particle PP moves in a horizontal plane. At time tt seconds its position vector relative to a fixed origin OO is r=(2t3−3t)i+(t2+4t)j\mathbf{r}=(2t^3-3t)\mathbf{i}+(t^2+4t)\mathbf{j} metres, where i\mathbf{i} and j\mathbf{j} are perpendicular unit vectors.
    Find the velocity of PP when t=2t=2.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).