Variable acceleration and vectors with calculus flashcards

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Velocity (motion in a straight line)
The rate of change of displacement with respect to time: v = dx/dt.
Acceleration (motion in a straight line)
The rate of change of velocity with respect to time: a = dv/dt = d²x/dt².
Instantaneously at rest
The velocity is zero at that instant (v = 0). If v changes sign there, the particle changes direction.
Displacement vs distance travelled
Displacement is the position relative to O (can be negative); distance travelled counts all movement, forwards and backwards, and is never negative.
Position vector r
The vector from a fixed origin O to the particle, r = xi + yj, with x and y functions of t.
Speed (from a velocity vector)
The magnitude of the velocity: for v = pi + qj, speed = √(p² + q²).
Velocity and acceleration from x(t)
v = dx/dt, a = dv/dt = d²x/dt². (No — learn it)
Velocity and displacement from a(t)
v = ∫a dt + c, x = ∫v dt + k; constants from initial conditions. (No — learn it)
Vector calculus for motion in a plane
v = dr/dt, a = dv/dt; r = ∫v dt + c, v = ∫a dt + c (component by component). (No — learn it)
Condition for pi + qj to be parallel to αi + βj
pi + qj = k(αi + βj), i.e. βp = αq. k > 0: same direction; k < 0: opposite direction. (No — derive it)
Resultant force from a variable acceleration
F = ma, with a = dv/dt at the required time; |F| = √(Fx² + Fy²). (No — learn it)
Differentiating and integrating powers of t
d/dt(tⁿ) = ntⁿ⁻¹; ∫tⁿ dt = tⁿ⁺¹/(n + 1) + c for n ≠ −1. Works for negative and fractional n, e.g. ∫27t⁻² dt = −27t⁻¹ + c. (No — learn it)
On an x–t graph, what does the gradient represent, and where is the particle at rest?
The gradient is the velocity; the particle is at rest where the graph is flat (gradient zero).
On a v–t graph, what represents displacement and what represents distance?
Signed area between the curve and the t-axis = displacement; total area, counting parts below the axis as positive = distance.
On a v–t graph, where is the greatest velocity?
At a turning point where the gradient (acceleration) is zero, or at an end of the interval — compare them.
On a sketch of a particle's path in the plane, what direction is the velocity?
Along the tangent to the path, in the direction of travel.
Sketching the motion of a particle that turns round
Draw a number line with O, mark the positions at t = 0, at each time v = 0, and at the end; arrows show each stage of the motion.
Why can't you use v = u + at when a = 6t?
The constant-acceleration formulae come from integrating a constant a. Here a varies, so integrate a(t) itself.
Can a particle be at rest with non-zero acceleration?
Yes. When it turns round, v = 0 for an instant but a ≠ 0, which is what reverses the motion.
Why does ∫v dt sometimes not give the distance travelled?
Backward motion makes negative contributions that cancel forward motion. Split the interval where v = 0.
Why can you differentiate r = xi + yj one component at a time?
The unit vectors i and j are constant, so d/dt(xi + yj) = (dx/dt)i + (dy/dt)j.
Why does v = 12 − 48t⁻² (t ≥ 2) never reach 12?
48t⁻² > 0 for all t, so v < 12 at every time: the velocity approaches 12 as a limit.
Conditions for a particle to be moving due north
The i-component of v is zero and the j-component is positive.
x = 2t³ − 9t² + 12t. When is the particle at rest?
v = 6t² − 18t + 12 = 6(t − 1)(t − 2) = 0, so t = 1 and t = 2.
a = 4 − 2t and v = 5 when t = 0. Greatest velocity?
v = 5 + 4t − t²; dv/dt = 0 at t = 2, v = 5 + 8 − 4 = 9 m s⁻¹.
v = t² − 4t + 3, starting at O. Distance travelled for 0 ≤ t ≤ 3?
At rest at t = 1 and 3. x = t³/3 − 2t² + 3t: x(1) = 4/3, x(3) = 0. Distance = 4/3 + 4/3 = 8/3 m (displacement 0).
r = t²i + (t³ − 3t)j. Speed when t = 2?
v = 2ti + (3t² − 3)j = 4i + 9j at t = 2; speed = √97 ≈ 9.85 m s⁻¹.
a = 6ti + 2j and v = i − 4j when t = 0. Find v, and when the particle moves due east.
v = (3t² + 1)i + (2t − 4)j. j-component zero at t = 2, i-component 13 > 0, so due east at t = 2.
v = (t − 1)i + (2t − 5)j. When is v parallel to i + j?
t − 1 = 2t − 5, so t = 4; then v = 3i + 3j = 3(i + j).
A 2 kg particle has a = 2ti − 3j. Magnitude of the resultant force at t = 2?
a = 4i − 3j, F = 8i − 6j, |F| = √(64 + 36) = 10 N.
Greatest speed over an interval
Find stationary values of v (dv/dt = 0) and compare |v| there with |v| at both ends of the interval.
Distance travelled questions
Always solve v = 0 first; if v changes sign inside the interval, split the integral and add the sizes.
Integrating a vector
Add a constant vector each time; the first comes from the initial velocity, the second from the initial position.
Given a force, not an acceleration
Divide F by the mass first, then integrate a to get v.
Two roots for “due north” or “parallel to”
Reject negative times, roots where the other component has the wrong sign, and roots where v = 0 (no direction).
Motion defined in stages
Use each formula only on its own time interval; the velocity at the change-over time links the stages and is a useful check.

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