# calculus

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6. The forward motion of a space shuttle, t seconds after touchdown, is described by
s(t) = 189t − t^3 , where " is measured in metres.
a) What is the velocity of the shuttle at touchdown?
b) How much time is required for the shuttle to stop completely?
c) How far does the shuttle travel from touchdown to a complete stop?
d) What is the deceleration 8 seconds after touchdown?

• calculus -

d)
v = 189 - 3t^2
a = -6t
so when t = 8, a = .....

(how easy was that? )

• calculus -

s= 189t- t³
v=ds/dt = 189-3t²
t=0 = > v=189 m/s.

189-3t² = 0
t=√(189/3) =√63 = 7.94 s.
s=189 t- t³=
=189(7.94) -7.94³=1000.1 m

a=dv/dt =d(189-3t²)/dt= - 6t
t=8 => a= - 6•8 = - 48 m/s²

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