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 $$v = 6i - 5 \frac{3}{2} j$$ 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 b... 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 first differentiate the velocity vector with respect to time:

$a = \frac{dv}{dt}$

Given that:

$v = 6i - 5 \frac{3}{2} j$

Differentiating:

$a = \frac{d}{dt}(6i - \frac{15}{2}j)$

As there are no time-dependent factors in the expression, we have:

$a = 0i - 0j = 0$ m s².

At t = 4 seconds, the acceleration remains the same and is 0 m s².

Step 2

Find the position vector of P when t = 4.

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Answer

To find the position vector, we first need to integrate the velocity vector:

$r(t) = r_0 + \int_0^t v(t) \, dt$

Given the initial position vector:

$r(0) = (-20i + 20j)$ m,

Integrating the velocity function:

$r(t) = (-20i + 20j) + \int_0^t (6i - \frac{15}{2} j) dt$

Calculating the integral:

Substituting t = 4: $$r(4) = (-20i + 20j) + (6(4)i - \frac{15}{2}(4)j)$$ $$= (-20i + 20j) + (24i - 30j)$$ $$= (4i - 10j) m.$

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