A-Level Edexcel Maths Mechanics Working with Vectors: A particle, P, moves with consta

A particle, P, moves with constant acceleration (2i - 3j) m s² At time t = 0, the particle is at the point A and is moving with velocity (-i + 4j) m s⁻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)-m-s²-At-time-t-=-0,-the-particle-is-at-the-point-A-and-is-moving-with-velocity-(-i-+-4j)-m-s⁻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) m s² At time t = 0, the particle is at the point A and is moving with velocity (-i + 4j) m s⁻¹ At time t = ... show full transcript

Worked Solution & Example Answer:A particle, P, moves with constant acceleration (2i - 3j) m s² At time t = 0, the particle is at the point A and is moving with velocity (-i + 4j) m s⁻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.

96%

114 rated

Answer

To find the value of T, we will use the equation of motion for velocity:

$v = u + at$

where:

Initial velocity, $u = -i + 4j$ m/s
Acceleration, $a = (2i - 3j)$ m/s²
Letting $v = k(3i - 4j)$ for some scalar $k$ as P needs to be in the direction of the vector $(3i - 4j)$ at time $t = T$ .

Now, substituting into the equation:

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

Separating the components gives us two equations:

$3k = -1 + 2T$
$-4k = 4 - 3T$

From the first equation: $k = \frac{-1 + 2T}{3}$

Substituting for $k$ in the second equation:

$-4\left(\frac{-1 + 2T}{3}\right) = 4 - 3T$

Multiplying by 3 to eliminate the fraction: $-4(-1 + 2T) = 3(4 - 3T)$ $4 - 8T = 12 - 9T$

Bringing like terms together gives: $T = 8$

Step 2

Find the distance AB.

99%

104 rated

Answer

To find the distance AB, we first find the positions of points A and B:

Position of A at t = 0: Given: $u = -i + 4j$ m/s and $a = (2i - 3j)$ m/s². The position can be calculated by: $s_A = ut + \frac{1}{2}at^2$ At $t = 0$ : $s_A = 0$
Position of B at t = 4: Using: $s_B = u(4) + \frac{1}{2}(2i - 3j)(4^2)$ $= (-i + 4j)(4) + \frac{1}{2}(2i - 3j)(16)$ $= -4i + 16j + (16i - 24j)$ $= (12i - 8j)$ .

Now, the distance AB is given by: $AB = |s_B - s_A| = |(12i - 8j)| = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}$ .

So, the distance AB is approximately $14.42$ m.

Find the value of T.

Find the distance AB.

Join the A-Level students using SimpleStudy...