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?
To find the answers to the given questions, we'll need to use calculus. We'll go step by step to find each answer.
a) To find the velocity, we need to take the derivative of the position function s(t). The derivative of s(t) with respect to t gives us the rate of change of position, which is the velocity.
First, let's find the derivative of s(t):
ds(t)/dt = d/dt (189t - t^3)
Differentiating each term:
ds(t)/dt = 189(1) - 3t^2
Simplifying:
ds(t)/dt = 189 - 3t^2
To find the velocity at touchdown (t = 0), substitute t = 0 into the velocity function:
v(0) = 189 - 3(0)^2
v(0) = 189 m/s
Therefore, the velocity of the shuttle at touchdown is 189 m/s.
b) To find the time required for the shuttle to stop completely, we need to find the time when the velocity is zero. This occurs when the derivative of the position function, ds(t)/dt, is zero.
Setting ds(t)/dt = 0:
189 - 3t^2 = 0
Rearranging the equation:
3t^2 = 189
Dividing both sides by 3:
t^2 = 63
Taking the square root of both sides:
t = ±√63
Since we are looking for a positive time value, t = √63 ≈ 7.937.
Therefore, it takes approximately 7.937 seconds for the shuttle to stop completely.
c) To find the distance traveled from touchdown to a complete stop, we need to find the total displacement. We'll integrate the velocity function from t = 0 to t = 7.937:
∫[0, 7.937] (189 - 3t^2) dt
Evaluating the integral:
[189t - t^3/3] from 0 to 7.937
Substituting the upper and lower limits:
[(189 * 7.937) - (7.937^3/3)] - [(189 * 0) - (0^3/3)]
Simplifying:
[(189 * 7.937) - (7.937^3/3)] - 0
Calculating:
(1499.293 - 94.387) - 0 ≈ 1404.906
Therefore, the shuttle travels approximately 1404.906 meters from touchdown to a complete stop.
d) To find the deceleration 8 seconds after touchdown, we need to find the derivative of the velocity function. The derivative of the velocity function with respect to t gives us the rate of change of velocity, which is the acceleration or deceleration.
First, let's find the derivative of v(t):
dv(t)/dt = d/dt (189 - 3t^2)
Differentiating each term:
dv(t)/dt = 0 - 6t
Simplifying:
dv(t)/dt = -6t
To find the deceleration at 8 seconds after touchdown (t = 8), substitute t = 8 into the deceleration function:
a(8) = -6(8)
a(8) = -48 m/s^2
Therefore, the deceleration 8 seconds after touchdown is -48 m/s^2.