# 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 = โ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๐ฅ with respect to ๐ฅ is ๐ cos of ๐๐ฅ, where ๐ 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๐ฅ with respect to ๐ฅ is negative ๐ sin of ๐๐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 by six. So, we have two ๐ก is equal to negative ๐ 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 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.