A-Level Edexcel Maths Mechanics Constant Acceleration - 2D: A particle, P, moves with consta

A particle, P, moves with constant acceleration (2i - 3j) ms² At time t = 0, the particle is at the point A and is moving with velocity (-i + 4j) ms⁻¹ At time t = T seconds, P is moving in the direction of vector (3i - 4j) (a) Find the value of T - Edexcel - A-Level Maths Mechanics - Question 2 - 2019 - Paper 1

Question 2

A-particle,-P,-moves-with-constant-acceleration-(2i---3j)-ms²-At-time-t-=-0,-the-particle-is-at-the-point-A-and-is-moving-with-velocity-(-i-+-4j)-ms⁻¹-At-time-t-=-T-seconds,-P-is-moving-in-the-direction-of-vector-(3i---4j)--(a)-Find-the-value-of-T-Edexcel-A-Level Maths Mechanics-Question 2-2019-Paper 1.png

A particle, P, moves with constant acceleration (2i - 3j) ms² At time t = 0, the particle is at the point A and is moving with velocity (-i + 4j) ms⁻¹ At time t = T ... show full transcript

Worked Solution & Example Answer:A particle, P, moves with constant acceleration (2i - 3j) ms² At time t = 0, the particle is at the point A and is moving with velocity (-i + 4j) ms⁻¹ At time t = T seconds, P is moving in the direction of vector (3i - 4j) (a) Find the value of T - Edexcel - A-Level Maths Mechanics - Question 2 - 2019 - Paper 1

Step 1

Find the value of T.

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Answer

To find the time T, we can use the following equation:

$v = u + at$

where:

$v$ is the final velocity,
$u$ is the initial velocity,
$a$ is the acceleration, and
$t$ is the time.

Substituting the given values:

Initial velocity $u = -i + 4j$ m/s
Acceleration $a = 2i - 3j$ m/s²

At time $T$ , the velocity becomes:

$v = (-i + 4j) + (2i - 3j)T$

This simplifies to: $v = (-1 + 2T)i + (4 - 3T)j$

We also know that at time $T$ , P is moving in the direction of vector $(3i - 4j)$ . Therefore, we can express this as:

$(-1 + 2T, 4 - 3T) ext{ is in the direction of } (3, -4)$

Now, using the property of direction ratios:

$\frac{-1 + 2T}{3} = \frac{4 - 3T}{-4}$

Cross-multiplying gives:

$-4(-1 + 2T) = 3(4 - 3T)$

Expanding:

$4 - 8T = 12 - 9T$

Rearranging the equation:

$T = 8$

Step 2

Find the distance AB.

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Answer

To find the distance AB, we first need to determine the position of point B at time t = 4 seconds.

Using the equation for displacement:

$s = ut + \frac{1}{2}at^2$

where:

$s$ is the displacement,
$u = -i + 4j$ m/s,
$a = 2i - 3j$ m/s², and
$t = 4$ seconds.

Substituting these values into the equation:

$s = (-i + 4j)(4) + \frac{1}{2}(2i - 3j)(4^2)$

Calculating each term: $s = (-4i + 16j) + \frac{1}{2}(2i - 3j)(16)$ $= (-4i + 16j) + (16i - 24j)$

Now combining like terms: $s = (-4 + 16)i + (16 - 24)j$ $= 12i - 8j$ .

Thus, the coordinates of point B are (12, -8).

Now, we need to calculate the distance AB, which is given by the formula:

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

Assuming point A is at (0, 0), the coordinates of A are (0, 0) and B are (12, -8). Therefore:

$AB = \sqrt{(12 - 0)^2 + (-8 - 0)^2}$ $= \sqrt{12^2 + (-8)^2}$ $= \sqrt{144 + 64}$ $= \sqrt{208}$ $= 4\sqrt{13} ext{ m}$ .

Find the value of T.

Find the distance AB.

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