The position of an object given by $f(t)=5t^2-6t+13$ where $t$ is measured in seconds and the position is measured in meters. Find the position of the object when the velocity is 0.
I'm confused on this problem. Am I suppose to use derivatives on this problem? And where do I plug in my 0 at?
Please Help!!
On
Use them rules from mechanics.
Differential of distance -->Velocity(in m/s)
Differential of velocity-->Acceleration(m/s^2)
Differential of accelaration-->Jerk(m/s^3).
So differentiate distance w.r.t time to get velocity. then equate that to zero and solve for t. Take that time t and substitute it back into the distance equation to get the distance.
So ds/dt = 10t - 6
equate that to 0 making it 0=10t - 6 and solve for t to get something like 6/10 seconds. take that 6/10 and substitute it into the function given for distance which is f(x)=5t^2 -6t +13 to get the distance in metres
HINT: The velocity is the derivative of the position with respect to time, which is $f\,'(t)$. Once you have that function, you set it to $0$ and solve for $t$ to find the time at which the velocity is zero, and then you can use that time to find the position of the object at that moment.