Location, Velocity, Acceleration, Jerk, Snap, Crackle and Pop.

Here is another (very elementary) code solution. My notation denotes all quantities from the "zeroth derivative" to the "sixth derivative". The first derivative is the velocity, and the sixth derivative is the same as "pop".

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import math

xd0 = 0.0
xd1 = 0.0
xd2 = 0.0
xd3 = 0.0
xd4 = 0.0
xd5 = 0.0
xd6 = 0.001

t = 0.0
dt = 10.0**(-6.0)

while xd0 <= 997.5:

    xd0 = xd0 + xd1 * dt  # Explicit Euler
    xd1 = xd1 + xd2 * dt
    xd2 = xd2 + xd3 * dt
    xd3 = xd3 + xd4 * dt
    xd4 = xd4 + xd5 * dt
    xd5 = xd5 + xd6 * dt

    t = t + dt

print dt
print t
print xd1

#1e-06
#29.9254670091
#199.996912521

$\text{DSolve}\left[\left\{x^{(6)}(t)=p,x^{(5)}(0)=c,x^{(4)}(0)=s,x^{(3)}(0)=j,x''(0)=a,x'(0)=v,x(0)=l\right\},x,t\right] \Rightarrow \\ \text{position}: t\to l+v\,t+ \frac{a}{2} t^2+ \frac{j}{6} t^3+\frac{s}{24} t^4+\frac{c}{120} t^5+\frac{p}{720} t^6$

With only $\text{pop}$ non-zero and equal to $\frac{1}{1000} \frac{m}{s^6}$ , the formula simplifies to $\text{position}: t\to \frac{1}{720\times1000}t^6$ .

Solving $\frac{1}{720\times1000}t^6=997.5$ gives $t=29.9254644949264$ seconds.

Doing the first derivative of position with respect to time and substituting these parameters, including the solved for time gives an answer after rounding to the nearest integer of $200\frac{m}{s}$ . This is about 447.4 mph.