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A particle P moves with constant acceleration (2i - 5j) m s^-2 - Edexcel - A-Level Maths Mechanics - Question 1 - 2009 - Paper 1

Question 1

A particle P moves with constant acceleration (2i - 5j) m s^-2. At time t = 0, P has speed u m s^-1. At time t = 3 s, P has velocity (-6i + j) m s^-1. Find the valu... show full transcript

Worked Solution & Example Answer:A particle P moves with constant acceleration (2i - 5j) m s^-2 - Edexcel - A-Level Maths Mechanics - Question 1 - 2009 - Paper 1

Step 1

At time t = 3 s, P's velocity equation

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Answer

The velocity \( extbf{v} \) of a particle is given by the equation:

\[ extbf{v} = extbf{u} + extbf{a}t \ \]

Where:

\textbf{v} = (-6i + j) m s^-1
\textbf{u} = initial velocity (unknown, denoted as u)
\textbf{a} = constant acceleration (2i - 5j) m s^-2
t = 3 s

Substituting the values:

\[ -6i + j = extbf{u} + 3(2i - 5j) \]

Step 2

Solving for initial velocity u

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Answer

Expanding the right side:

\[ -6i + j = extbf{u} + (6i - 15j) \]

Rearranging gives:

\[ extbf{u} = -6i + j - (6i - 15j) \] \[ extbf{u} = -6i + j - 6i + 15j \] \[ extbf{u} = -12i + 16j \]

Step 3

Finding the speed u at t = 0

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Answer

To find the speed u, we calculate the magnitude of the velocity vector:

\[ u = || extbf{u}|| = \sqrt{(-12)^2 + (16)^2} \]

Calculating the components:

\[ u = \sqrt{144 + 256} \]

Thus:

\[ u = \sqrt{400} = 20 \]

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