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A particle moves along the π₯-axis such that at time π‘ seconds, its displacement from the origin is given by π₯ equals 10 sin of two π‘ meters, for π‘ is greater than or equal to zero.
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Determine the particleβs acceleration when π₯ is equal to negative five.
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Here, weβre given a function for displacement at time π‘ seconds.
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And weβre being asked to find the particleβs acceleration.
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Before we can evaluate the acceleration when π₯ is equal to negative five meters, weβre simply going to find expressions for the particleβs acceleration.
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And so, we begin by recalling that the velocity is change in displacement with respect to time.
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In terms of derivatives, we can say π£ is equal to dπ₯ by dπ‘.
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Similarly, acceleration is change in velocity with respect to time.
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So, in terms of derivatives, thatβs dπ£ by dπ‘.
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Since π£ is dπ₯ by dπ‘, we can say that acceleration could also be written as d two π₯ by dπ‘ squared.
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Now, weβre told that π₯ is equal to 10 sin of two π‘.
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Letβs differentiate this twice with respect to π‘ to find an expression for the acceleration.
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We know that the derivative of sin ππ₯ with respect to π₯ is π cos of ππ₯, where π is a real constant and π₯ is measured in radians.
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Since π₯ is 10 sin of two π‘, we say that dπ₯ by dπ‘ is 10 times two cos of two π‘, or 20 cos of two π‘.
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Weβll differentiate one more time to find our expression for acceleration.
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This time though, we know that the derivative of cos ππ₯ with respect to π₯ is negative π sin of ππ₯.
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So, the second derivative of π₯ with respect to π‘ is 20 times negative two sin of two π‘, which is negative 40 sin of two π‘.
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The question asks us to evaluate this when π₯ is equal to negative five.
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So, we need to work out the value of π‘ at this point.
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And so, we go back to our original equation.
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We let π₯ be equal to negative five, and weβre going to solve for π‘.
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We divide through by 10.
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And we know that negative five divided by 10 is negative one-half.
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So, our equation is negative one-half equals sin of two π‘.
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Then, we take the arc sin or inverse sin of both sides such that two π‘ is equal to the inverse sin of negative one-half.
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But the inverse sin of negative one-half is negative π by six.
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So, we have two π‘ is equal to negative π by six.
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We could use this to work out the value of π‘.
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But if we go back to our expression for acceleration, we see that weβre going to substitute in two π‘ anyway.
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And so, we find that the second derivative of π₯ with respect to π‘, which we know is acceleration, is negative 40 sin of negative π by six, which is negative 40 times negative one-half, or 20.
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The particleβs acceleration when π₯ is equal to negative five meters is 20 meters per square second.