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

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:

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