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?

d)

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

(how easy was that? )

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

To find the answers to these questions, we need to apply the concepts of velocity, stopping time, distance, and deceleration.

a) The velocity of an object is defined as the rate of change of displacement with respect to time. In this case, the displacement function is given as s(t) = 189t − t^3. To find the velocity, we need to calculate the derivative of s(t) with respect to t.

Taking the derivative of s(t):

s'(t) = d/dt (189t − t^3)
= 189 − 3t^2

Therefore, the velocity of the shuttle at touchdown (t = 0) is s'(0) = 189 − 3(0)^2 = 189 m/s.

b) To find the time required for the shuttle to stop completely, we need to solve for t when the velocity (s'(t)) becomes zero.

0 = 189 − 3t^2

Rearranging this equation:

3t^2 = 189

Dividing by 3:

t^2 = 63

Taking the square root of both sides:

t = ± sqrt(63)

Since time cannot be negative, the time required for the shuttle to stop completely is t ≈ 7.937 seconds.

c) To determine the distance traveled by the shuttle from touchdown to a complete stop, we need to calculate the integral of the velocity function over the time interval [0, t]. Recall that the integral of velocity gives us the displacement.

The integral of s'(t) with respect to t:

s(t) = ∫(189 − 3t^2) dt
= 189t − t^3/3 + C

Evaluating the integral from 0 to t:

s(t) = [(189t − t^3/3) - (189(0) − (0)^3/3)]
= 189t − t^3/3

Substituting t ≈ 7.937:

s(7.937) = 189(7.937) − (7.937)^3/3 ≈ 750.543 meters

Therefore, the shuttle travels approximately 750.543 meters from touchdown to a complete stop.

d) To find the deceleration 8 seconds after touchdown, we need to calculate the derivative of the velocity function s'(t).

Taking the derivative of s'(t):

s''(t) = d/dt (189 − 3t^2)
= -6t

Therefore, the deceleration 8 seconds after touchdown (t = 8) is s''(8) = -6(8) = -48 m/s^2.