A particle moves so that its acceleration, a ms⁻², at time t seconds may be modelled in terms of its velocity, v ms⁻¹, as a = -0.1v² The initial velocity of the particle is 4 ms⁻¹. 19 (a) By first forming a suitable differential equation, show that v = \frac{20}{5 + 2t} 19 (b) Find the acceleration of the particle when t = 5.5.

A-Level AQA Maths Pure Differential Equations: A particle moves so that its acc

A particle moves so that its acceleration, a ms⁻², at time t seconds may be modelled in terms of its velocity, v ms⁻¹, as a = -0.1v² The initial velocity of the particle is 4 ms⁻¹ - AQA - A-Level Maths Pure - Question 19 - 2020 - Paper 2

Question 19

A particle moves so that its acceleration, a ms⁻², at time t seconds may be modelled in terms of its velocity, v ms⁻¹, as a = -0.1v² The initial velocity of the part... show full transcript

Worked Solution & Example Answer:A particle moves so that its acceleration, a ms⁻², at time t seconds may be modelled in terms of its velocity, v ms⁻¹, as a = -0.1v² The initial velocity of the particle is 4 ms⁻¹ - AQA - A-Level Maths Pure - Question 19 - 2020 - Paper 2

Step 1

By first forming a suitable differential equation, show that

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Answer

To form the differential equation, we start from the relationship between acceleration and velocity:

$a = \frac{dv}{dt}$

Substituting the given expression for acceleration:

$\frac{dv}{dt} = -0.1v^2$

Next, we separate the variables for integration:

$\frac{dv}{v^2} = -0.1 dt$

Now, we integrate both sides. The left side integrates to:

$-\frac{1}{v} = -0.1t + C$

Rearranging gives:

$\frac{1}{v} = 0.1t + C$

To find the constant C, we use the initial condition that when t = 0, v = 4:

$\frac{1}{4} = 0 + C \Rightarrow C = \frac{1}{4}$

Thus, we have:

$\frac{1}{v} = 0.1t + \frac{1}{4}$

Inverting gives:

$v = \frac{1}{0.1t + \frac{1}{4}} = \frac{1}{\frac{1}{4} + 0.1t} = \frac{4}{1 + 0.4t}$

To express this in the required form:

$v = \frac{20}{5 + 2t}$

Step 2

Find the acceleration of the particle when t = 5.5.

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Answer

We previously established the relationship between velocity and time. First, we need to find the velocity at t = 5.5:

$v = \frac{20}{5 + 2(5.5)} = \frac{20}{5 + 11} = \frac{20}{16} = 1.25 \ ext{ms}⁻¹$

Next, we can find the acceleration using the equation for acceleration:

$a = -0.1v^2$

Substituting the value of v:

$a = -0.1(1.25)^2 = -0.1(1.5625) = -0.15625 \ ext{ms}⁻²$

Thus, the acceleration of the particle when t = 5.5 is approximately -0.1563 ms⁻².

A particle moves so that its acceleration, a ms⁻², at time t seconds may be modelled in terms of its velocity, v ms⁻¹, as a = -0.1v² The initial velocity of the particle is 4 ms⁻¹ - AQA - A-Level Maths Pure - Question 19 - 2020 - Paper 2

Worked Solution & Example Answer:A particle moves so that its acceleration, a ms⁻², at time t seconds may be modelled in terms of its velocity, v ms⁻¹, as a = -0.1v² The initial velocity of the particle is 4 ms⁻¹ - AQA - A-Level Maths Pure - Question 19 - 2020 - Paper 2

By first forming a suitable differential equation, show that

Find the acceleration of the particle when t = 5.5.

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