What is the Velocity Function for a Car Slowing at a Uniform Rate?

[Discalculiate]

The position of a car that is being slowed at a uniform rate is given by:
s(t) = 25t - 5.25t^2
where s is the displacement of the car from the point at which braking began and t is the
time in seconds after braking began. Find the velocity function for the car, including
its domain.

The answer is v(t) = 25 -11t 0 < or equal to t < or equal to 25/11

I assume I need to find the value of t first, then take a derivative, but I'm not sure
how to get t by itself
0 = 25t - 5.25t^2
5.25t^2 = 25t
and then I don't know what to do.

 

[harshey]

I may be wrong here, someone correct me if I am.

I think since s(t) is the position of the car, simply differentiating the equation will give you the v(t), velocity of the car with respect to time.

Thus, v(t) = 25-10.5t

I'm not sure how the answer is v(t) = 25-11t
Maybe the original problem was written incorrectly and should have been s(t)=25t-5.5t^2?

Hopefully I helped...

 

[Discalculiate]

Ohh, thanks!
I think maybe he might have rounded? That's probably it, but I don't know why.

so if v(t) = 25 - 11t
then when the velocity is 0, the car has come to a stop so
0 = 25 - 11t
11t = 25
so t = 25/11

 

[harshey]

yup. no problem.