At time $t$ seconds, a particle $P$ has velocity $ extbf{v} ext{ m s}^{-1}$, where $$ extbf{v} = 3t^{2} extbf{i} - 2t extbf{j} \, (t > 0)$$ (a) Find the acceleration of $P$ at time $t$ seconds, where $t > 0$ - Edexcel - A-Level Maths Mechanics - Question 5 - 2021 - Paper 1

For $P$ to move in the direction of $extbf{i} - extbf{j}$, we require:

$extbf{v} ext{ is parallel to } extbf{i} - extbf{j}$

This means:

rac{3t^{2}}{-2t} = rac{1}{-1}

Solving this:

$3t^{2} = 2t$

So:

\implies t = 0 ext{ or } t = \frac{2}{3}$$ Thus, at the instant required, $t = \frac{2}{3}$.

The position vector $r$ is given by integrating the velocity:

$r = \int extbf{v} \, dt = \int (3t^{2} extbf{i} - 2t extbf{j}) \, dt$

This gives:

$r = t^{3} extbf{i} - t^{2} extbf{j} + C$

Using the initial condition when $t=1$, $r = - extbf{j}$, we can find $C$:

$1^{3} extbf{i} - 1^{2} extbf{j} + C = - extbf{j}$

This implies:

$C = - extbf{i} - extbf{j}$

Thus, the expression becomes:

$r = t^{3} extbf{i} - t^{2} extbf{j} - extbf{i} - extbf{j} = (t^{3} - 1) extbf{i} - (t^{2} + 1) extbf{j}$

First, find the speed:

$| extbf{v}| = ext{speed} = \sqrt{(3t^{2})^{2} + (-2t)^{2}}$

Setting this equal to 10:

$\sqrt{9t^{4} + 4t^{2}} = 10$

Squaring both sides:

$9t^{4} + 4t^{2} = 100$

Letting $u = t^{2}$, we have:

$9u^{2} + 4u - 100 = 0$

Using the quadratic formula:

$u = \frac{-4 \pm \sqrt{4^{2} - 4 \cdot 9 \cdot -100}}{2 \cdot 9}$

Calculating:

After solving, we find:

$u = 4$\nso $t=2$.

Next, find $r$ at $t=2$:

$r = (2^{3} - 1) extbf{i} - (2^{2} + 1) extbf{j} = 7 extbf{i} - 5 extbf{j}$

The distance is:

$|r| = \sqrt{7^{2} + (-5)^{2}} = \sqrt{49 + 25} = \sqrt{74}$