Lesson 6.06a
6.06a Motion under variable forces and differential equations of motion Quiz: OCR Further Maths, Unit 3
20 questions
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Lesson 6.06a, Motion under variable forces and differential equations of motion: 20 multiple choice questions for the OCR Further Maths (H245), Unit 3: Mechanics (Y543), written with Revision Ninja.
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The 20 questions
-
Which expression for acceleration should be used when velocity is given as a function of displacement x?
- v dx/dt
- d^2x/dt^2
- v dv/dx
- dv/dt
-
Which expression for acceleration is used when velocity is explicitly given as a function of time t?
- dv/dt
- d^2t/dx^2
- v dx/dt
- v dv/dx
-
A particle of mass 2 kg moves under a force F = 6t. What is its acceleration at t = 3?
- 3 m s^-2
- 9 m s^-2
- 36 m s^-2
- 18 m s^-2
-
What method solves a differential equation of the form dv/dt = g(t)h(v)?
- Separation of variables
- Partial fractions
- Integrating factor
- Euler's method
-
What is the integrating factor for the first-order linear differential equation dv/dt + P(t)v = Q(t)?
- e^(-∫ P dt)
- ∫ P dt
- e^(∫ Q dt)
- e^(∫ P dt)
-
If v dv/dx = 4x and v = 0 at x = 0, what is v in terms of x?
- 2x
- 2x^2
- 4x
- x^2
-
A particle moves with acceleration a = -kv. What type of differential equation models this motion?
- Second-order linear
- Exact quadratic
- Non-linear homogeneous
- First-order linear
-
If acceleration a = 3v, which separable differential equation determines displacement x in terms of v?
- v dv/dx = 3
- dv/dx = 3v
- dx/dv = 3v
- dv/dx = 3
-
A force F = 8 - 2v acts on a 1 kg particle. What is its terminal velocity?
- 8 m s^-1
- 2 m s^-1
- 4 m s^-1
- 0 m s^-1
-
Integrating dv/dt = 10 - 2v using an integrating factor requires rewriting it in which standard form?
- dv/dt - 2v = 10
- dv/dt + 2v = 10
- 2v dv/dt = 10
- dv/dt + 10v = 2
-
What is the velocity v(t) if dv/dt = -2v with initial velocity v(0) = 5?
- 5 e^(2t)
- 2 e^(-5t)
- 5 e^(-2t)
- 5 - 2t
-
For motion under a variable force F(x), what does Newton's second law state?
- m d^2x/dv^2 = F(x)
- m dv/dx = F(x)
- m v dv/dt = F(x)
- m v dv/dx = F(x)
-
A particle has acceleration a = 6t - 2. If v(0) = 3, what is v(2)?
- 14 m s^-1
- 8 m s^-1
- 10 m s^-1
- 11 m s^-1
-
What is the integrating factor for the differential equation dv/dt + 3t^2 v = t^2?
- t^3
- e^(t^3)
- e^(3t^2)
- e^(3t)
-
If a = v^2 and v(0) = 1, what is v in terms of x when x(0) = 0?
- 1 + x
- e^(2x)
- 1/(1-x)
- e^x
-
What condition defines the maximum velocity of a particle moving in a straight line under a variable force?
- Force is maximum
- Velocity is zero
- Acceleration is zero
- Displacement is zero
-
If a particle of mass 3 kg has acceleration a = 4x, what is the force acting on it?
- 12x
- 4x/3
- 12x^2
- 4x
-
Which equation links displacement x and velocity v when acceleration is a function of velocity, f(v)?
- dx/dv = 1 / f(v)
- dx/dv = f(v) / v
- dx/dv = f(v)
- dx/dv = v / f(v)
-
Solve v dv/dx = 9 given v = 0 at x = 0. What is v at x = 2?
- 3
- 18
- 36
- 6
-
If dv/dt + v = e^(-t) and v(0) = 0, what is the integrating factor?
- e^(2t)
- e^t
- e^(-t)
- t
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