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A cyclist, moving in a straight line, accelerated over a distance of 35.5 meters until his velocity reached 10.8 meters per second.
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Given that this took five seconds, find the cyclistβs initial velocity.
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We can answer this question using our equations of uniform acceleration known as the SUVAT equations.
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π indicates the displacement, π’ the initial velocity, π£ the final velocity, π the acceleration, and π‘ the time.
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We are told that the cyclist covers a distance of 35.5 meters.
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This means that his displacement is 35.5.
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He reaches a velocity of 10.8 meters per second.
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So this is our value of π£.
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As the time taken was five seconds, π‘ is equal to five.
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We need to calculate the initial velocity π’.
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We will use the equation π is equal to π’ plus π£ divided by two multiplied by π‘.
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Substituting in our values, we have 35.5 is equal to π’ plus 10.8 divided by two all multiplied by five.
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We can divide both sides of this equation by five such that 7.1 is equal to π’ plus 10.8 divided by two.
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Multiplying by two, we have 14.2 is equal to π’ plus 10.8.
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Finally, subtracting 10.8 from both sides of this equation gives us π’ is equal to 3.4.
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The cyclistβs initial velocity is 3.4 meters per second.