A-Level Edexcel Maths Mechanics Forces: At time t seconds, where t

At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v m s⁻¹ is given by $$ extbf{v} = t extbf{i} - 4 extbf{j}$$ When t = 1, P is at the point A and when t = 4, P is at the point B - Edexcel - A-Level Maths Mechanics - Question 6 - 2018 - Paper 2

Question 6

At-time-t-seconds,-where-t->-0,-a-particle-P-moves-in-the-x-y-plane-in-such-a-way-that-its-velocity-v-m-s⁻¹-is-given-by--$$-extbf{v}-=-t--extbf{i}---4--extbf{j}$$--When-t-=-1,-P-is-at-the-point-A-and-when-t-=-4,-P-is-at-the-point-B-Edexcel-A-Level Maths Mechanics-Question 6-2018-Paper 2.png

At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v m s⁻¹ is given by $$ extbf{v} = t extbf{i} - 4 extbf{j}$$ W... show full transcript

Worked Solution & Example Answer:At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v m s⁻¹ is given by $$ extbf{v} = t extbf{i} - 4 extbf{j}$$ When t = 1, P is at the point A and when t = 4, P is at the point B - Edexcel - A-Level Maths Mechanics - Question 6 - 2018 - Paper 2

Step 1

Find position at time t = 1 (Point A)

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Answer

To find the position of particle P at time t = 1, we integrate the velocity function. The position function ( extbf{s}(t) ) is found by integrating the velocity:

$extbf{s}(t) = \int \textbf{v} \ dt = \int (t \textbf{i} - 4 \textbf{j}) \ dt = \left( \frac{t^2}{2} \textbf{i} - 4t \textbf{j} \right) + extbf{C}$

We take the initial conditions at t = 1 to find C. Set ( \textbf{s}(1) = \textbf{A} ):

$\textbf{s}(1) = \left( \frac{1^2}{2} \textbf{i} - 4*1 \textbf{j} \right) + \textbf{C} = \frac{1}{2} \textbf{i} - 4\textbf{j} + \textbf{C}$

Now we can deduce that ( \mathbf{A} ) represents the position at t = 1.

Step 2

Find position at time t = 4 (Point B)

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Similarly, we calculate the position vector at t = 4:

$\textbf{s}(4) = \left( \frac{4^2}{2} \textbf{i} - 4*4 \textbf{j} \right) + \textbf{C} = (8 \textbf{i} - 16 \textbf{j}) + \textbf{C}$

Using the position determined previously at t = 1, we find that ( \textbf{C} ) remains unchanged, and thus we calculate the position at t = 4, giving us point B.

Step 3

Calculate the distance AB

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Answer

With the positions defined:

Point A is given by ( \frac{1}{2} \textbf{i} - 4 \textbf{j} )
Point B is given by ( 8 \textbf{i} - 16 \textbf{j} )

The distance ( AB ) is calculated as follows:

$AB = \| \textbf{B} - \textbf{A} \| = \sqrt{\left( 8 - \frac{1}{2} \right)^2 + \left( -16 - (-4) \right)^2}$

Which simplifies to:

$AB = \sqrt{\left( \frac{15}{2} \right)^2 + (-12)^2} = \sqrt{\frac{225}{4} + 144} = \sqrt{ \frac{225 + 576}{4} } = \sqrt{ \frac{801}{4} } = \frac{\sqrt{801}}{2}$

Thus, the exact distance AB is ( \frac{\sqrt{801}}{2} ).

At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v m s⁻¹ is given by $$ extbf{v} = t extbf{i} - 4 extbf{j}$$ When t = 1, P is at the point A and when t = 4, P is at the point B - Edexcel - A-Level Maths Mechanics - Question 6 - 2018 - Paper 2

Find position at time t = 1 (Point A)

Find position at time t = 4 (Point B)

Calculate the distance AB

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