### Video Transcript

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

Here, weโre given a function for displacement at time ๐ก seconds. And weโre being asked to find the particleโs acceleration. Before we can evaluate the acceleration when ๐ฅ is equal to negative five meters, weโre simply going to find expressions for the particleโs 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 ๐ฃ is equal to d๐ฅ by d๐ก. Similarly, acceleration is change in velocity with respect to time. So, in terms of derivatives, thatโs d๐ฃ by d๐ก. Since ๐ฃ is d๐ฅ by d๐ก, we can say that acceleration could also be written as d two ๐ฅ by d๐ก squared.

Now, weโre told that ๐ฅ is equal to 10 sin of two ๐ก. Letโs differentiate this twice with respect to ๐ก to find an expression for the acceleration. We know that the derivative of sin ๐๐ฅ with respect to ๐ฅ is ๐ cos of ๐๐ฅ, where ๐ is a real constant and ๐ฅ is measured in radians. Since ๐ฅ is 10 sin of two ๐ก, we say that d๐ฅ by d๐ก is 10 times two cos of two ๐ก, or 20 cos of two ๐ก.

Weโll differentiate one more time to find our expression for acceleration. This time though, we know that the derivative of cos ๐๐ฅ with respect to ๐ฅ is negative ๐ sin of ๐๐ฅ. So, the second derivative of ๐ฅ with respect to ๐ก is 20 times negative two sin of two ๐ก, which is negative 40 sin of two ๐ก.

The question asks us to evaluate this when ๐ฅ is equal to negative five. So, we need to work out the value of ๐ก at this point. And so, we go back to our original equation. We let ๐ฅ be equal to negative five, and weโre going to solve for ๐ก. 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 ๐ก. Then, we take the arc sin or inverse sin of both sides such that two ๐ก is equal to the inverse sin of negative one-half. But the inverse sin of negative one-half is negative ๐ by six. So, we have two ๐ก is equal to negative ๐ by six.

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