A particle moves on a straight line with a constant acceleration, a m s⁻². The initial velocity of the particle is U m s⁻¹. After T seconds the particle has velocity V m s⁻¹. This information is shown on the velocity-time graph. The displacement, S metres, of the particle from its initial position at time T seconds is given by the formula S = \frac{1}{2}(U + V)T 12 (a) By considering the gradient of the graph, or otherwise, write down a formula for a in terms of U, V and T. 12 (b) Hence show that V² = U² + 2aS.

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A particle moves on a straight line with a constant acceleration, a m s⁻² - AQA - A-Level Maths Mechanics - Question 12 - 2017 - Paper 2

Question 12

A particle moves on a straight line with a constant acceleration, a m s⁻². The initial velocity of the particle is U m s⁻¹. After T seconds the particle has velocity... show full transcript

Worked Solution & Example Answer:A particle moves on a straight line with a constant acceleration, a m s⁻² - AQA - A-Level Maths Mechanics - Question 12 - 2017 - Paper 2

Step 1

By considering the gradient of the graph, or otherwise, write down a formula for a in terms of U, V and T.

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Answer

To find the formula for acceleration, a, we can use the relationship between the initial and final velocities and the time taken. The gradient of the velocity-time graph gives acceleration, which can be expressed as:

$a = \frac{V - U}{T}$

Step 2

Hence show that V² = U² + 2aS.

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Answer

Starting from the formula for displacement:

$S = \frac{1}{2}(U + V)T$

We can rearrange it to express T:

$T = \frac{V - U}{a}$

Substituting for T back into the displacement formula, we have:

Substitute for T: $S = \frac{1}{2}(U + V) \left( \frac{V - U}{a} \right)$ Thus, $2as = (U + V)(V - U)$
Expanding this and rearranging gives: $2as = V^2 - U^2$
Therefore, $V^2 = U^2 + 2aS$

This completes the required proof.

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