(a) Given that m = i - i, n = 2i + 3j and 2m + n - r = 0, find |r| (b) The distance, S metres of a moving particle at any time tseconds is given by S = 3t -...

(a) Given that m = i - i, n = 2i + 3j and 2m + n - r = 0, find |r|

(b) The distance, S metres of a moving particle at any time tseconds is given by

S = 3t - \(\frac{t^3}{3}\) + 9

Find the;

(i) time

(ii) distance travelled

When the particle is momentarily at rest

Answer Details

(a) Given that m = i - i, m = 0. Also, n = 2i + 3j.

Since 2m + n = 2(0) + n = n = 2i + 3j, the equation becomes:

2m + n - r = 2(0) + n - r = n - r = 0

Therefore, r = 2i + 3j.

The magnitude of a vector can be found using the Pythagorean theorem, which states that the magnitude of a vector (a, b) is given by √(a^2 + b^2).

So, |r| = √(2^2 + 3^2) = √(4 + 9) = √13.

(b) (i) The particle is momentarily at rest when its velocity is 0, i.e., when its derivative with respect to time is 0. The velocity of the particle is given by the derivative of its position with respect to time.

So, the velocity is given by:

v = dS/dt = 3 - t^2

Setting this equal to 0 and solving for t, we get:

0 = 3 - t^2

t^2 = 3

t = ±√3

So, the particle is momentarily at rest when t = ±√3.

(ii) To find the distance travelled when the particle is momentarily at rest, we need to substitute t = ±√3 into the expression for S:

S = 3t - t^3/3 + 9

S = 3(±√3) - (±√3)^3/3 + 9

S = 9 ± 3√3 - (3 ± 3√3) + 9

S = 18

So, the particle has travelled a distance of 18 metres when it is momentarily at rest.