A particle moves along the 𝑥-axis such that at time 𝑡 seconds, its displacement from the origin is given by 𝑥 equals 10 sin of two 𝑡 meters, for 𝑡 is greater than or equal to zero. Determine the particle’s acceleration when 𝑥 is equal to negative five.
Here, we’re given a function for displacement at time 𝑡 seconds. And we’re being asked to find the particle’s acceleration. Before we can evaluate the acceleration when 𝑥 is equal to negative five meters, we’re simply going to find expressions for the particle’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 𝑣 is equal to d𝑥 by d𝑡. Similarly, acceleration is change in velocity with respect to time. So, in terms of derivatives, that’s d𝑣 by d𝑡. Since 𝑣 is d𝑥 by d𝑡, we can say that acceleration could also be written as d two 𝑥 by d𝑡 squared.
Now, we’re told that 𝑥 is equal to 10 sin of two 𝑡. Let’s differentiate this twice with respect to 𝑡 to find an expression for the acceleration. We know that the derivative of sin 𝑎𝑥 with respect to 𝑥 is 𝑎 cos of 𝑎𝑥, where 𝑎 is a real constant and 𝑥 is measured in radians. Since 𝑥 is 10 sin of two 𝑡, we say that d𝑥 by d𝑡 is 10 times two cos of two 𝑡, or 20 cos of two 𝑡.
We’ll differentiate one more time to find our expression for acceleration. This time though, we know that the derivative of cos 𝑎𝑥 with respect to 𝑥 is negative 𝑎 sin of 𝑎𝑥. So, the second derivative of 𝑥 with respect to 𝑡 is 20 times negative two sin of two 𝑡, which is negative 40 sin of two 𝑡.
The question asks us to evaluate this when 𝑥 is equal to negative five. So, we need to work out the value of 𝑡 at this point. And so, we go back to our original equation. We let 𝑥 be equal to negative five, and we’re going to solve for 𝑡. 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 𝑡. Then, we take the arc sin or inverse sin of both sides such that two 𝑡 is equal to the inverse sin of negative one-half. But the inverse sin of negative one-half is negative 𝜋 by six. So, we have two 𝑡 is equal to negative 𝜋 by six.
We could use this to work out the value of 𝑡. But if we go back to our expression for acceleration, we see that we’re going to substitute in two 𝑡 anyway. And so, we find that the second derivative of 𝑥 with respect to 𝑡, which we know is acceleration, is negative 40 sin of negative 𝜋 by six, which is negative 40 times negative one-half, or 20. The particle’s acceleration when 𝑥 is equal to negative five meters is 20 meters per square second.