suppose that the position of a particle as a function of time is given by the formula s(t)= 10+2t^7-t^9,t>0
the time at which the velocity is zero is
the time at which the acceleration is zero is
the time at which the velocity is maximum is
65Given function is `s(t)=10+2t^7-t^9` . Here `t` is denoting time .
Now we know that velocity of a body is the rate of change of dispalcement. In calculus we write velocity
Now we need to determine the time at which velocity is zero i.e. `d/dt(s)=0` .
So, either t^6=0 or 14-9t^2=0
Because t>0, so the first condition t^6=0 is not possible. It means 14-9t^2=0
or, t=sqrt(14/9)=+sqrt(14)/3. Answer.
Acceleration means rate of change of velocity i.e. dv/dt.
The time at which acceleration is zero means we find the value of t by equating dv/dt=0
Because t>0. So 7-6t^2=0
or, t=+sqrt(7/6) . Answer.
For maximum or minimum of the velocity function dv/dt=0
i.e. acceleration is zero. i.e. t=+sqrt(7/6)
Now at t=sqrt(7/6)
So the velocity of the particle will be maximum at t=sqrt(7/6). Answer.
The position is given by `s(t)=10+2t^7-t^9` for t>0.
The velocity function is the first derivative of the position function. The acceleration is the derivative of the velocity function (or the second derivative of the position function.) The velocity function takes on its maximum or minimum when the acceleration function is zero.
(a) The velocity is zero when `14t^6-9t^8=0`
`t^6(14-9t^2)=0==> t^6= 0,(14-9t^2)=0`
since t>0 we consider `14-9t^2=0 ==> t^2=14/9`
Again since t>0 we have `t=sqrt(14)/3~~1.25`
The velocity is zero when t is approximately 1.25
(b) The acceleration is zero when `84t^5-72t^7=0`
`12t^5(7-6t^2)=0 ==> 12t^5=0,7-6t^2=0`
Since t>0 we have `t^2=7/6==>t=sqrt(7/6)~~1.08`
The acceleration is zero at t approximately 1.08
(c) The velocity is maximum when the acceleration is zero. For positive t, this occurs when t is approximately 1.08 and the velocity is approximately 5.56
The graph of the position function in black, the velocity function in red, and the acceleration in green: