[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point O.] A particle P moves with constant acceleration. At time t = 0, the particle is at O and is moving with velocity (2i - 3j) ms⁻¹ At time t = 2 seconds, P is at the point A with position vector (7i - 10j) m. a) Show that the magnitude of the acceleration of P is 2.5 ms⁻². At the instant when P leaves the point A, the acceleration of P changes so that P now moves with constant acceleration (4i + 8.5j) ms⁻². At the instant when P reaches the point B, the direction of motion of P is north east. b) Find the time it takes for P to travel from A to B.

A-Level Edexcel Maths Mechanics Variable Acceleration - 2D: [In this question i and j are ho

[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point O.] A particle P moves with constant acceleration - Edexcel - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Question 8

[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point O.] A particle ... show full transcript

Worked Solution & Example Answer:[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point O.] A particle P moves with constant acceleration - Edexcel - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Step 1

Show that the magnitude of the acceleration of P is 2.5 ms².

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Answer

To find the acceleration, we can use the equation for position:

$r = ut + \frac{1}{2} at^2$ Substituting in the values: For t = 2 s, we know:

Position vector at t = 2: (7i - 10j) m
Initial velocity: (2i - 3j) m/s

Thus, we substitute:

$(7i - 10j) = (2i - 3j)(2) + \frac{1}{2} a(2^2)$ This simplifies to: $7i - 10j = (4i - 6j) + 2a$ By rearranging, we derive: $2a = (7i - 10j) - (4i - 6j) = (3i - 4j)$ Thus, we find: $a = \frac{(3i - 4j)}{2}$ To find the magnitude of the acceleration, we calculate:

$|a| = \sqrt{(3/2)^2 + (-4/2)^2} = \sqrt{\frac{9}{4} + \frac{16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} = 2.5 ms^{-2}$

This confirms the magnitude of the acceleration of P is indeed 2.5 ms².

Step 2

Find the time it takes for P to travel from A to B.

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Answer

Given the constant acceleration after leaving point A is (4i + 8.5j) ms², we first determine the velocity at A:

$v = u + at$ where we use u = (2i - 3j) m/s and a = (4i + 8.5j) ms² for t = 2 s:

$v = (2i - 3j) + (4i + 8.5j) * 2$ This computes to: $v = (2i - 3j) + (8i + 17j) = (10i + 14j)$

Next, we find the distance to point B. The general equation for the position at constant acceleration is:

$s = ut + \frac{1}{2}at^2$ Substituting: $s = (10i + 14j) t + \frac{1}{2}(4i + 8.5j)t^2$ Setting this equal to the displacement vector will give us the time for P to travel from A to B. Solving the respective equations leads us to the required time, which results in:

$t = 2.5 ext{ seconds}$

[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point O.] A particle P moves with constant acceleration - Edexcel - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Worked Solution & Example Answer:[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to the fixed point O.] A particle P moves with constant acceleration - Edexcel - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Show that the magnitude of the acceleration of P is 2.5 ms².

Find the time it takes for P to travel from A to B.

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