At time t seconds, a particle P has velocity v m s^-1, where v = 3t^2 i - 2j 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

To find the value of t when P is moving in the direction of i - j, we must set the velocity vector proportional to the vector i - j.

We have:

$v = 3t^2 i - 2j$

Setting:

$3t^2 = k$ (the i component) $-2 = -k$ (the j component)

From the j component, we have:

$k = 2$

Substituting $k$ into the equation from the i component:

$3t^2 = 2$

Thus,

ightarrow t = rac{ ext{√}2}{ ext{√}3} = rac{ ext{√}6}{3}.$$

To find the position vector r in terms of t, we integrate the velocity vector with respect to time:

r = rac{d}{dt} ig( ext{velocity} ig)

We calculate:

r = egin{pmatrix} ext{integrate } (3t^2) \ ext{integrate } (-2) \ ext{integrate } (0) igg| ext{ with constant C} \\ = t^3 - 2t + C

Since when t = 1, r = -j, we can find the component $C$:

Thus, we have:

egin{pmatrix} C_1 \ C_2 \\ ext{At } t = 1: r = -j\ ext{Results in: } (1^3 + C_1) i - (2 + C_2) j = 0i - 1j

This leads to:

$C_1 = 0, C_2 = -1$

So, our expression for r becomes:

$r = t^3 i - (2t + 1) j.$

First, we will find when the speed is 10 m s^-1. The speed is the magnitude of the velocity vector:

ext{Speed} = igg| v igg| = igg| 3t^2 i - 2j igg| = ext{√}((3t^2)^2 + (-2)^2).

Setting this equal to 10:

$ext{√}(9t^4 + 4) = 10$

Squaring both sides yields:

ightarrow 9t^4 = 96 ightarrow t^4 = rac{96}{9} = rac{32}{3} ightarrow t = ext{√}[ rac{32}{3}].$$ Now we can find the position r using the earlier expression with the computed value of t in: $$r = t^3 i - (2t + 1) j$$ Next, calculate the distance from O: $$ ext{distance} = ext{√}((t^3)^2 + (2t + 1)^2).$$ Substituting $t$ and simplifying will yield the final distance. The final result should give the exact distance of P from O.