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 Mechanics Variable Acceleration - 1D: 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 Mechanics - 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 pa... 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 Mechanics - Question 19 - 2020 - Paper 2

Step 1

By first forming a suitable differential equation, show that v = \frac{20}{5 + 2t}

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Answer

To derive the relationship between velocity and time, we start from the given acceleration:

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

We can rearrange this to separate variables. Dividing both sides by $v^{2}$ and multiplying by $dt$ gives:

$\frac{1}{v^{2}} dv = -0.1 dt$

Next, we integrate both sides:

$\int \frac{1}{v^{2}} dv = \int -0.1 dt$

The left side integrates to $-\frac{1}{v}$ , and the right side integrates to $-0.1t + C$ . Thus, we have:

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

Multiplying throughout by -1 yields:

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

Now, we can express v as:

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

To determine the constant $C$ , we can use the initial condition where at $t = 0$ , $v = 4$ :

$4 = \frac{1}{-C} \implies C = -\frac{1}{4}$

Now substituting this back into the equation gives:

$v = \frac{1}{0.1t + \frac{1}{4}} = \frac{1}{\frac{5}{4} + 0.1t} = \frac{20}{5 + 2t}$

Thus, we have shown that:

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

Step 2

Find the acceleration of the particle when t = 5.5.

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Answer

To find the acceleration when $t = 5.5$ , we first need to find the velocity at that time using the equation derived above:

$v = \frac{20}{5 + 2(5.5)} = \frac{20}{5 + 11} = \frac{20}{16} = 1.25 \text{ ms}^{-1}$

Next, we substitute this value of $v$ back into the acceleration formula:

$a = -0.1v^{2} = -0.1(1.25)^{2} = -0.1 \times 1.5625 = -0.15625 \text{ ms}^{-2}$

So, the acceleration of the particle when $t = 5.5$ seconds is approximately:

$a \approx -0.1563 \text{ ms}^{-2}$

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 Mechanics - 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 Mechanics - Question 19 - 2020 - Paper 2

By first forming a suitable differential equation, show that v = \frac{20}{5 + 2t}

Find the acceleration of the particle when t = 5.5.

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