Question

a particles velocity, in \\(\frac{m}{s}\\), at a particular time \\(t\\) is given by the function \\(v(t) = \sqrt{2t^5 + 2t^3 + 6}\\). what is the particles velocity, in \\(\frac{m}{s}\\), at \\(t = 9.7\\) seconds? round to one decimal place.
show your work here
hint: to add an exponent (\\(x^y\\)), type \exponent\ or press \^\

Explanation:

Step1: Substitute t = 9.7 into the function

We need to calculate \(2t^{5}+2t^{3}+6\) when \(t = 9.7\). First, calculate \(t^{5}\) and \(t^{3}\):
\(t^{3}=(9.7)^{3}=9.7\times9.7\times9.7 = 912.673\)
\(t^{5}=(9.7)^{5}=(9.7)^{2}\times(9.7)^{3}=94.09\times912.673\approx85877.47\)
Then, calculate \(2t^{5}\) and \(2t^{3}\):
\(2t^{5}=2\times85877.47 = 171754.94\)
\(2t^{3}=2\times912.673 = 1825.346\)
Now, add them with 6:
\(2t^{5}+2t^{3}+6=171754.94 + 1825.346+6=173586.286\)

Step2: Take the square root

Now, we need to find the square root of \(173586.286\):
\(v(9.7)=\sqrt{173586.286}\approx416.6\) (rounded to one decimal place)

Answer:

\(416.6\)