Help with Integration/Kinematics Question

[Cherries]

A driver of a car travelling on a straight slippery road sees a stop sign 50 m ahead and starts to break. After the breaks are applied, the velocity of the car is given by v(t)=11-t^(3/2). Will the car go beyond the stop sign?

I'm specifically having trouble with determining the C value for the indefinite equation of displacement.

Thanks!

[kw0573]

Both definite and indefinite integrals can be used to solve this problem. I prefer the definite integral slightly, because it does not make an arbitrary choice. 

The indefinite integral ∫ v(t) dt gives the displacement function, which describes the distance from a reference point, as a function of time. An arbitrary, yet useful, decision is to make the displacement function pass through the origin. In other words, we use the location at which brakes were first applied as the reference point. d(t) = ∫ v(t) dt = 11t - (2/5)*t^(5/2) + C, where C is zero. We know that v(t) = 0 when t = 11^(2/3). Plug in this stopping time into d(t) we find that d(11^(2/3)) = 32.6 < 50. It takes the driver 32.6 m to stop, so it is before the stop sign.

The definite integral can remove this arbitrary choice. ∫ v(t) dt from t = 0 to t = 11^(2/3) gives the expression d(11^(2/3)) - d(0). This means that when the driver speed changes during t = 0 and t = 11^(2/3), the net displacement is the change in the position of those times. Without defining what C is, we can find ∫ v(t) dt from t = 0 to t = 11^(2/3) = 32.6 < 50.

Many other solutions are possible, such as finding t that makes d(t) = 50 and whether it is greater than or less than 11^(2/3).