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A particle P moves with constant acceleration (2i - 3j) ms² - Edexcel - A-Level Maths Mechanics - Question 1 - 2021 - Paper 1
Question 1
A particle P moves with constant acceleration (2i - 3j) ms². At time t = 0, P is moving with velocity 4 ms⁻¹. (a) Find the velocity of P at time t = 2 seconds. At ... show full transcript
Worked Solution & Example Answer:A particle P moves with constant acceleration (2i - 3j) ms² - Edexcel - A-Level Maths Mechanics - Question 1 - 2021 - Paper 1
Step 1
Find the velocity of P at time t = 2 seconds.
Answer
To find the velocity of the particle at time t, we use the formula: v = u + at where:
- u = initial velocity = 4i m/s
- a = acceleration = (2i - 3j) m/s²
- t = time = 2 seconds
Substituting the values in:
v = (4i) + (2i - 3j) * 2$$ Calculating the product of acceleration and time:v = 4i + (4i - 6j)$$
v = 8i - 6j$$ Thus, the velocity of P at time t = 2 seconds is given by:v = 8i - 6j ext{ m/s}$$
Step 2
Find the position vector of P relative to O at time t = 3 seconds.
Answer
First, we find the velocity at time t = 3 using the previously calculated velocity:
u = 8i - 6j$$ Now, we compute the position vector, using: $$r = r_0 + vt + \frac{1}{2}at^2$$ where: - r₀ = initial position vector = (i + j) m = (1i + 1j) m - v = (8i - 6j) m/s - a = (2i - 3j) m/s² - t = 3 seconds Substituting into the equation: $$r = (i + j) + (8i - 6j) * 3 + \frac{1}{2} (2i - 3j) * (3^2)$$ Calculating each term: 1. The displacement from the velocity term: $$= (8i - 6j) * 3 = 24i - 18j$$ 2. The displacement from the acceleration term: $$= \frac{1}{2} (2i - 3j) * 9 = (9i - 13.5j)$$ Now summing them all together: $$r = (1 + 24 + 9)i + (1 - 18 - 13.5)j$$ $$r = 34i - 30.5j$$ Thus, the position vector of P at t = 3 seconds is: $$r = 34i - 30.5j ext{ m}$$