Answer:
1) s(3) = -32 feet.
2)v(5) = 3 feet/sec
3)a(4) = 12[tex]feet/s^{2}[/tex]
4) Velocity becomes zero at t = 5 seconds
Explanation:
Given that position as a function of time is
[tex]s(t)=t^{3}-6t^{2}-15t+40[/tex]
Now by definition of velocity we have
[tex]v=\frac{ds}{dt}\\\\v=\frac{d}{dt}(t^{3}-6t^{2}-15t+40)\\\\\therefore v(t)=3t^{2}-12t-15[/tex]
Now by definition of acceleration we have
[tex]a=\frac{dv}{dt}\\\\a=\frac{d}{dt}(3t^{2}-12t-15)\\\\\therefore a(t)=6t-12[/tex]
Applying values of time in corresponding equations we get
1) s(3)=[tex]3^{3}-6\times (3)^{2}-15\times 3+40=-32feet[/tex]
2)v(5)=[tex]3\times {5}^{2}-12\times 5-15=3feet/sec[/tex]
3)a(4)=[tex]6\times 4-12=12ft/s^{2}[/tex]
4)To obatin the time at which velocity is zero equate the velocity function with zero we get
[tex]3t^{2}-12t-15=0\\\\t^{2}-4t-5=0\\\\t^{2}-5t+t-5=0\\\\t(t-5)+1(t-5)=0\\\\(t-5)(t+1)=0\\\\\therefore t=5\\\\or\\\therefore t=-1[/tex]
Thus the correct time is 5 seconds at which velocity becomes zero.
