Answer: 1) Instantaneous velocity at t= 8 equals 81m/s
2) Instantaneous acceleration at t= 8 equals [tex]36m/s^{2}[/tex]
3) Average velocity between 0 to 8 seconds equals 1 m/s.
4) Average speed between 0 to 8 seconds equals 23.31 m/s.
Explanation:
Given position as a function of time as
[tex]s(t)=t^3-6t^2-15t+7[/tex]
Now by definition of velocity 'v' we have
[tex]v=\frac{d}{dt}\cdot s(t)\\\\v=\frac{d}{dt}\cdot (t^3-6t^2-15t+7)\\\\v=3t^2-12t-15[/tex]
Thus velocity at t = 8 seconds equals [tex]v(8)=3\times 8^2-12\times 8-15=81m/s[/tex]
Now by definition of acceleration 'a' we have
[tex]a=\frac{d}{dt}\cdot v(t)\\\\a=\frac{d}{dt}\cdot (3t^2-12t-15)\\\\a=6t-12[/tex]
Thus acceleration at t = 8 seconds equals [tex]a(8)=6\times 8-12=36m/s^{2}[/tex]
Part 2)
The average of any function is mathematically defined as
[tex]\overline{v}=\frac{\int v(t)dt}{\int dt}[/tex]
Using the given function of velocity and using the limits from t = 0 to t = 8 secs we get
[tex]\overline{v}=\frac{\int_{0}^{8}(3t^2-12t-15)}{\int_{0}^{8}dt}\\\\\therefore \overline{v}=\frac{8}{8}=1m/s[/tex]
the average speed is calculated as
[tex]Speed=\frac{Distance}{Time}[/tex]
[tex]Distance=\int \sqrt{1+(\frac{ds}{dt})^{2}}\cdot dt\\\\Distance=\int_{0}^{8}\sqrt{1+(3t^2-12t-15)^{2}}\cdot dt\\\\Distance=186.51m[/tex]
Hence the average speed is calculated as
[tex]Speed=\frac{186.51}{8}=23.31m/s[/tex]
