At time t seconds, a particle P has velocity v m s^-1, where v = 3t^2 i - 2j 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 be moving in the direction of the vector i - j, we need the velocity vector to be proportional to it. Setting up the equation:

$v = 3t^2 i - 2j = k(i - j)$ for some scalar k, we equate components:

1. 3t^2 = k
2. -2 = -k

From the second equation, we get k = 2. Substituting this into the first gives:

$3t^2 = 2\Rightarrow t^2 = \frac{2}{3}\Rightarrow t = \sqrt{\frac{2}{3}}.$ Therefore, the value of t is:

$t = \frac{\sqrt{6}}{3}.$

The position vector r can be found by integrating the velocity v with respect to time:

$r = \int v \ dt = \int (3t^2 i - 2j) \ dt = (t^3) i - (2t) j + C.$ Using the condition at t = 1 where r = -j:

When t = 1, $r = (1^3)i - (2 * 1)j + C = i - 2j + C.$ We know that this equals -j, hence: $i - 2j + C = -j \Rightarrow C = -i.$ Thus, $r = (t^3 - 1)i - 2j.$

First, we must find the time when the speed is 10 m s^-1. The speed is given by:

$|v| = \sqrt{(3t^2)^2 + (-2)^2} = \sqrt{9t^4 + 4}.$ Setting this equal to 10:

$\sqrt{9t^4 + 4} = 10 \Rightarrow 9t^4 + 4 = 100 \Rightarrow 9t^4 = 96 \Rightarrow t^4 = \frac{32}{3} \Rightarrow t = \left(\frac{32}{3}\right)^{1/4}.$ Now, substituting this value into the expression for r, we find:

$r = \left(\left(\frac{32}{3}\right)^{1/4^3} - 1\right)i - 2\left(\frac{32}{3}\right)^{1/4}j.$ To find the distance from O, we calculate:

$|r| = \sqrt{(\left(\frac{32}{3}\right)^{1/4^3} - 1)^2 + (-2\left(\frac{32}{3}\right)^{1/4})^2}.$ And simplifying this expression will yield the exact distance.