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?
To find the answers to these questions, we need to understand the concepts of velocity, time, and deceleration.
a) Velocity is the rate of change of displacement with respect to time. The velocity at touchdown is the derivative of the position function s(t) with respect to time, t. So, to find the velocity, we need to take the derivative of the given position function.
s(t) = 189t - t^3
Taking the derivative of s(t) with respect to t, we get:
v(t) = s'(t) = d/dt (189t - t^3)
= 189 - 3t^2
Therefore, the velocity of the shuttle at touchdown is v(0) = 189 - 3(0)^2 = 189 m/s.
b) To find the time required for the shuttle to stop completely, we need to find the value of t when the velocity v(t) becomes zero. So, we need to set the velocity function equal to zero and solve for t.
v(t) = 189 - 3t^2
0 = 189 - 3t^2
3t^2 = 189
t^2 = 63
t = sqrt(63)
Therefore, the time required for the shuttle to stop completely is t = sqrt(63) seconds.
c) The distance traveled by the shuttle from touchdown to a complete stop is the total displacement during this interval. To find this, we need to integrate the velocity function from t = 0 to t = sqrt(63).
s(t) = ∫(189 - 3t^2) dt (from 0 to sqrt(63))
= 189t - t^3/3 | from 0 to sqrt(63)
= 189(sqrt(63)) - (sqrt(63))^3/3 - (0 - 0)
Therefore, the distance traveled by the shuttle from touchdown to a complete stop is 189(sqrt(63)) - (sqrt(63))^3/3 meters.
d) Deceleration is the negative rate of change of velocity with respect to time. To find the deceleration 8 seconds after touchdown, we need to take the derivative of the velocity function with respect to time.
a(t) = v'(t) = d/dt (189 - 3t^2)
= -6t
Therefore, the deceleration 8 seconds after touchdown is a(8) = -6(8) = -48 m/s^2.