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 = 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 V^2 = U^2 + 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 veloc... 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

The gradient of a velocity-time graph represents acceleration. Thus, acceleration, a, can be expressed as:

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

Step 2

Hence show that V^2 = U^2 + 2aS.

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Answer

From the formula for displacement, we have:

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

We can rearrange this formula to express T in terms of S, U, and V:

$T = \frac{2S}{U + V}$
Now substituting T in the formula for acceleration:

$a = \frac{V - U}{T} = \frac{V - U}{\frac{2S}{U + V}} = \frac{(V - U)(U + V)}{2S}$
Rearranging gives:

$2aS = (V - U)(U + V)$
Expanding the right-hand side leads us to:

$2aS = V^2 - U^2$
Finally, rearranging this equation yields:

$V^2 = U^2 + 2aS$

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