[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 Working with Vectors: [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 ms$^{-2}$

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Answer

To find the acceleration of the particle P, we first utilize the kinematic equation for position:

$r = u t + \frac{1}{2} a t^2$

Where:

r is the position vector,
u is the initial velocity,
a is the acceleration,
t is the time.

Substituting the known values, at time t = 2:

r = (7i - 10j) m,
u = (2i - 3j) ms $^{-1}$ ,

We can express the equation as follows:

$(7i - 10j) = (2i - 3j)(2) + \frac{1}{2}a(2^2)$

Calculating the left side gives:

$(7i - 10j) = (4i - 6j) + 2a$

Rearranging the equation to isolate a results in:

$2a = (7i - 10j) - (4i - 6j)$

This simplifies to:

$2a = (3i - 4j)$

From which we find:

$a = \frac{1}{2}(3i - 4j) = (1.5i - 2j) ms^{-2}$

Now to calculate the magnitude of the acceleration:

$|a| = \sqrt{(1.5)^2 + (-2)^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5 ms^{-2}$

Step 2

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

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Answer

At the instant P leaves point A, the velocity can be found using the equation:

$v = u + at$

Where:

u is the velocity at A,
a is the acceleration vector (4i + 8.5j) ms $^{-2}$ ,
t is the time taken to move from A to B.

We know:

u = (5i - 7j) as determined just before it reaches A,

The equation becomes:

$v = (5i - 7j) + (4i + 8.5j)t$

At the point B, the velocity vector must have a direction that is northeast. Therefore, we can write:

$v_y / v_x = 1$

Thus, by equating the components, we can solve for t:

For the components in i we have:

$v_x = 5 + 4t$

For the components in j:

$v_y = -7 + 8.5t$

Setting these equal gives us:

$-7 + 8.5t = 5 + 4t$

Rearranging to solve for t results in:

$4.5t = 12 \Rightarrow t = \frac{12}{4.5} = 2.67 s \approx 2.67 s$

[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 ms$^{-2}$

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

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