A-Level Edexcel Maths Mechanics Newtons Second Law: 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 $$v = t^2 i - 4j$$ 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--$$v-=-t^2-i---4j$$--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 $$v = t^2 i - 4j$$ When t = 1, P is at the ... 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 $$v = t^2 i - 4j$$ 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 the position of point A when t = 1

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Answer

To find the position of point A, we need to integrate the velocity function with respect to time.

The velocity is given by:

$v = t^2 i - 4j$

Integrating this from 0 to 1:

$s = \int (t^2 i - 4j) dt = \left( \frac{t^3}{3} i - 4t j \right) + C$

Assuming P starts at the origin (0,0) when t=0:

At t = 0, s = 0:

ightarrow C = 0$$ Now plugging in t = 1: $$s(1) = \left( \frac{1^3}{3} i - 4(1) j \right) = \left( \frac{1}{3} i - 4j \right)$$ So, point A is at \( A \left( \frac{1}{3}, -4 \right) \).

Step 2

Find the position of point B when t = 4

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Answer

Using the same integral result:

For t = 4:

$s(4) = \left( \frac{4^3}{3} i - 4(4) j \right) = \left( \frac{64}{3} i - 16j \right)$

So, point B is at ( B \left( \frac{64}{3}, -16 \right) ).

Step 3

Calculate the distance AB

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Answer

The distance AB can be computed using the distance formula:

$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$

Where:

( x_A = \frac{1}{3} )
( y_A = -4 )
( x_B = \frac{64}{3} )
( y_B = -16 )

Substituting these values in:

$AB = \sqrt{\left( \frac{64}{3} - \frac{1}{3} \right)^2 + (-16 + 4)^2}$

This simplifies to:

$AB = \sqrt{\left( \frac{63}{3} \right)^2 + (-12)^2}$

Calculating further:

$AB = \sqrt{21^2 + 12^2} = \sqrt{441 + 144} = \sqrt{585} = 3\sqrt{65}$

Thus, the exact distance AB is ( 3\sqrt{65} ).

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 $$v = t^2 i - 4j$$ 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 the position of point A when t = 1

Find the position of point B when t = 4

Calculate the distance AB

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