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A moving particle is defined by the two equations π₯ equals π‘ cubed minus five π‘ minus five and π¦ equals seven π‘ squared minus three.
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Find the magnitude of the acceleration of the particle at π‘ equals one.
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In this question, weβve been given the position of the particle as defined by a pair of parametric equations.
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So given a value of π‘, we obtain a coordinate pair π₯π¦ for the position of our particle.
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We could choose to consider this in vector terms and say that the position of the particle at time π‘, π of π‘, is given by π‘ cubed minus five π‘ minus five π plus seven π‘ squared minus three π.
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Now, in this question, weβre looking to find the acceleration.
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Well, in fact, we want the magnitude of the acceleration, but weβll deal with that in a moment.
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So we recall that the acceleration is equal to the derivative of the velocity with respect to time.
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But we also know that the velocity is equal to the first derivative of the displacement or the position vector.
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We can in turn say that, to find a function for acceleration, weβre going to need to differentiate our function for position twice with respect to time.
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Weβll do it once to find the function for velocity.
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We can differentiate each component function in turn.
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When we differentiate π‘ cubed minus five π‘ minus five, we get three π‘ squared minus five.
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And when we differentiate seven π‘ squared minus three, we get 14π‘.
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So our function for velocity is three π‘ squared minus five π plus 14π‘ π.
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But what does this actually mean?
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Well, it means that the velocity can be defined in terms of its horizontal velocity and its vertical velocity.
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Horizontally, its velocity is three π‘ squared minus five, but vertically, itβs given by the function 14π‘.
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Okay, great.
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Letβs differentiate again to find our function for acceleration.
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And when we differentiate three π‘ squared minus five, we find that the acceleration in the horizontal direction is six π‘.
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We then differentiate 14π‘.
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And we see that the acceleration in the vertical direction is 14.
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Weβre now able to find a vector acceleration for our particle at π‘ equals one.
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We simply substitute π‘ equals one into a vector function for acceleration.
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And we find that the acceleration at π‘ equals one is given by the vector six π plus 14π.
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We, of course, want to find the magnitude of the acceleration though.
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So we recall that the magnitude of a vector in two dimensions given by π₯ π plus π¦π is the square root of π₯ squared plus π¦ squared.
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This means the magnitude of our acceleration at π‘ equals one is the square root of six squared plus 14 squared, which is two root 58.
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There are no units here.
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So we found the magnitude of the acceleration of the particle at π‘ equals one to be two root 58.