A particle moves along the 𝑥-axis. Its position from the origin is 𝑠 meters at a time 𝑡 seconds. The position is given by the equation 𝑠 is equal to two sin of five 𝜋 over seven 𝑡 plus seven. Find the times 𝑡 at which the particle’s velocity is equal to zero.
In this question, we will begin by finding an expression for the particle’s velocity at time 𝑡. We are told that the displacement from the origin at time 𝑡 is equal to two sin of five 𝜋 over seven 𝑡 plus seven. We recall that 𝑣 is equal to d𝑠 by d𝑡. We can differentiate our expression for the displacement 𝑠 with respect to 𝑡 to find an expression for the velocity 𝑣. Our expression for 𝑠 contains the sine function. And we know that if 𝑦 is equal to sin of 𝑎𝑥, then d𝑦 by d𝑥 is equal to 𝑎 multiplied by cos of 𝑎𝑥.
This means that in order to differentiate the first term, we begin by multiplying five 𝜋 over seven by two. Differentiating two sin of five 𝜋 over seven 𝑡 with respect to 𝑡 gives us 10𝜋 over seven cos of five 𝜋 over seven 𝑡. Differentiating a constant gives us zero. So this is our expression for d𝑠 by d𝑡 or the velocity 𝑣.
We are asked to find the times 𝑡 at which the velocity is equal to zero. In order to do this, we set the expression 10𝜋 over seven multiplied by cos of five 𝜋 over seven 𝑡 equal to zero. We can divide both sides by 10𝜋 over seven such that cos of five 𝜋 over seven 𝑡 equals zero. Next, we recall that if cos 𝜃 equals zero, 𝜃 is equal to 𝜋 over two plus 𝑛𝜋 where 𝑛 is an integer. In our question, cos of five 𝜋 over seven 𝑡 equals zero. And this means that five 𝜋 over seven 𝑡 must be equal to 𝜋 over two plus 𝑛𝜋.
We can simplify this equation by dividing through by 𝜋. This means that five-sevenths 𝑡 is equal to one-half plus 𝑛. We can then divide through by five-sevenths, which is the same as multiplying by seven-fifths giving us 𝑡 is equal to seven-tenths plus seven-fifths 𝑛. The times 𝑡 at which the particle’s velocity is equal to zero are 𝑡 is equal to seven-tenths plus seven-fifths 𝑛, where 𝑛 is an integer.