I've been doing math.

See, I want to find out what distance the Firebolt will travel in the first 10 seconds if it can go from 0 - 150 mph (0 - about 67 m/s) in 10 seconds. After a lot of calculations, I figured since I don't have the form of the equation that the vt graph is in, I can't do it. *wail* I really want to know the answer! I tried v = a^t - 1 (where v is velocity in m/s, t is time in s, and a is a constant), but I couldn't integrate somenumber^t... I tried v = e^at - 1 (v is velocity, t is time, a is a constant), but I got distance = about 150m which I think is quite unreasonable because the Firebolt is supposed to be fast and 150m in 10 seconds = about 54km/h and that's not fast at all.

Okay okay I show working k:

Taking 1 mile = 1609 metres,

max. speed (if max. speed is 150 mph)
= (150 X 1609) m/h
= 241350 m/h
= (241350/3600) m/s
= 67.04 m/s (4sf)

IF velocity of Firebolt can be expressed as v = e^(ax + b) - 1,
[An exponential graph makes sense to show the way the Firebolt would fly, and I need -1 in the exponential equation because the graph should pass through the origin where t=0, v=0, or it won't make sense.]

When t=0, v=0,

0 = e^(a(0) + b) - 1
e^b = 1
b = 0

Therefore equation is now v = e^ax - 1

When t=10, v=67.04,

67.04 = e^(a(10)) - 1
e^10a = 68.04
10a = ln68.04
a = 0.4220 (4sf)

Therefore equation is now v = e^(0.4220(x)) - 1

So if I want to find distance, s, I integrate that equation.
[Let's make { the integration sign thing.]

s = { e^(0.4220(t)) - 1 dt
s = e^(0.4220(t))/0.4220 - t + c

When t=0, s=0,

0 = 1/0.4220 - 0 + c
c = - 1/0.4220

Therefore,

s = e^(0.4220(t))/0.4220 - t - 1/0.4220

When t=10,

s = e^(0.4220(10))/0.4220 - 10 - 1/0.4220
s = 14.8471192 m

Which doesn't look right. =(

Someone tell me what I'm doing wrong!