In this question position vectors are given relative to a fixed origin O. At time t seconds, where t > 0, a particle, P, moves so that its velocity v ms⁻¹ is given by v = 6i - 5j When t = 0, the position vector of P is (-20i + 20j) m. (a) Find the acceleration of P when t = 4 (b) Find the position vector of P when t = 4

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In this question position vectors are given relative to a fixed origin O - Edexcel - A-Level Maths Mechanics - Question 1 - 2019 - Paper 1

Question 1

In this question position vectors are given relative to a fixed origin O. At time t seconds, where t > 0, a particle, P, moves so that its velocity v ms⁻¹ is given ... show full transcript

Worked Solution & Example Answer:In this question position vectors are given relative to a fixed origin O - Edexcel - A-Level Maths Mechanics - Question 1 - 2019 - Paper 1

Step 1

Find the acceleration of P when t = 4

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Answer

To find the acceleration, we differentiate the velocity vector with respect to time.

Given: $v = 6i - 5j$

We differentiate:

$a = \frac{dv}{dt} = \left( \frac{d(6)}{dt} i + \frac{d(-5)}{dt} j \right) = 0i + 0j = 0$

The acceleration of P at any time t, including when t = 4, is therefore:

$a = 0i + 0j \; (m \, s^{-2})$

Step 2

Find the position vector of P when t = 4

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Answer

To find the position vector, we need to integrate the velocity vector over time.

Using: $r(t) = r_0 + \int v dt$

We have the initial position vector when t = 0: $r_0 = (-20i + 20j) m$

Now, integrating the velocity:

$r(t) = r_0 + \int(6i - 5j) dt = r_0 + (6t i - 5t j) + C$

Calculating when t = 4: $r(4) = (-20i + 20j) + (6*4 i - 5*4 j)$ $= (-20i + 20j) + (24i - 20j)$ $= (-20 + 24)i + (20 - 20)j$ $= 4i + 0j = 4i \, (m)$

Thus, the position vector of P when t = 4 is:

$r(4) = 4i + 0j$

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