Question Video: Finding the Acceleration of a Body given Its Displacement as a Function of Time Mathematics

A particle moves along the ๐‘ฅ-axis such that at time ๐‘ก seconds its displacement from the origin is given by ๐‘ฅ = (10 sin 2๐‘ก) m, ๐‘ก โ‰ฅ 0. Determine the particleโ€™s acceleration when ๐‘ฅ = โˆ’5 m.

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Video Transcript

A particle moves along the ๐‘ฅ-axis such that at time ๐‘ก seconds, its displacement from the origin is given by ๐‘ฅ equals 10 sin of two ๐‘ก meters, for ๐‘ก is greater than or equal to zero. Determine the particleโ€™s acceleration when ๐‘ฅ is equal to negative five.

Here, weโ€™re given a function for displacement at time ๐‘ก seconds. And weโ€™re being asked to find the particleโ€™s acceleration. Before we can evaluate the acceleration when ๐‘ฅ is equal to negative five meters, weโ€™re simply going to find expressions for the particleโ€™s acceleration. And so, we begin by recalling that the velocity is change in displacement with respect to time. In terms of derivatives, we can say ๐‘ฃ is equal to d๐‘ฅ by d๐‘ก. Similarly, acceleration is change in velocity with respect to time. So, in terms of derivatives, thatโ€™s d๐‘ฃ by d๐‘ก. Since ๐‘ฃ is d๐‘ฅ by d๐‘ก, we can say that acceleration could also be written as d two ๐‘ฅ by d๐‘ก squared.

Now, weโ€™re told that ๐‘ฅ is equal to 10 sin of two ๐‘ก. Letโ€™s differentiate this twice with respect to ๐‘ก to find an expression for the acceleration. We know that the derivative of sin ๐‘Ž๐‘ฅ with respect to ๐‘ฅ is ๐‘Ž cos of ๐‘Ž๐‘ฅ, where ๐‘Ž is a real constant and ๐‘ฅ is measured in radians. Since ๐‘ฅ is 10 sin of two ๐‘ก, we say that d๐‘ฅ by d๐‘ก is 10 times two cos of two ๐‘ก, or 20 cos of two ๐‘ก.

Weโ€™ll differentiate one more time to find our expression for acceleration. This time though, we know that the derivative of cos ๐‘Ž๐‘ฅ with respect to ๐‘ฅ is negative ๐‘Ž sin of ๐‘Ž๐‘ฅ. So, the second derivative of ๐‘ฅ with respect to ๐‘ก is 20 times negative two sin of two ๐‘ก, which is negative 40 sin of two ๐‘ก.

The question asks us to evaluate this when ๐‘ฅ is equal to negative five. So, we need to work out the value of ๐‘ก at this point. And so, we go back to our original equation. We let ๐‘ฅ be equal to negative five, and weโ€™re going to solve for ๐‘ก. We divide through by 10. And we know that negative five divided by 10 is negative one-half. So, our equation is negative one-half equals sin of two ๐‘ก. Then, we take the arc sin or inverse sin of both sides such that two ๐‘ก is equal to the inverse sin of negative one-half. But the inverse sin of negative one-half is negative ๐œ‹ by six. So, we have two ๐‘ก is equal to negative ๐œ‹ by six.

We could use this to work out the value of ๐‘ก. But if we go back to our expression for acceleration, we see that weโ€™re going to substitute in two ๐‘ก anyway. And so, we find that the second derivative of ๐‘ฅ with respect to ๐‘ก, which we know is acceleration, is negative 40 sin of negative ๐œ‹ by six, which is negative 40 times negative one-half, or 20. The particleโ€™s acceleration when ๐‘ฅ is equal to negative five meters is 20 meters per square second.

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