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 analyze the equation given for the forward motion of the space shuttle. Let's break it down step by step:
a) What is the velocity of the shuttle at touchdown?
To find the velocity, we need to take the derivative of the position equation s(t) with respect to time, t. The derivative of s(t) gives us the velocity, v(t).
s(t) = 189t - t^3�
To find v(t), take the derivative of s(t) with respect to t:
v(t) = d(s(t))/dt
Taking the derivative of each term in the equation, we get:
v(t) = 189 - 3t^2
So, the velocity of the shuttle at touchdown (t=0) is given by:
v(0) = 189 - 3(0)^2
v(0) = 189 - 0
v(0) = 189 meters per second
Therefore, the velocity of the shuttle at touchdown is 189 meters per second.
b) How much time is required for the shuttle to stop completely?
To find the time it takes for the shuttle to stop completely, we set the velocity equation v(t) equal to zero and solve for t:
0 = 189 - 3t^2
Rearranging the equation, we get:
3t^2 = 189
Dividing both sides by 3, we have:
t^2 = 63
Taking the square root of both sides, we get:
t = √63 ≈ 7.94 seconds
Therefore, the time required for the shuttle to stop completely is approximately 7.94 seconds.
c) How far does the shuttle travel from touchdown to a complete stop?
To find the distance traveled, we need to find the integral of the velocity equation v(t) over the time interval from touchdown (t=0) to when the shuttle stops (t=7.94):
∫[0 to 7.94] (189 - 3t^2) dt
Evaluating the integral, we get:
s(t) = 189t - t^3/3 (evaluated from 0 to 7.94)
Plugging in the values, we get:
s(7.94) = 189(7.94) - (7.94)^3/3
s(7.94) ≈ 1497.7 - 33.3
s(7.94) ≈ 1464.4 meters
Therefore, the shuttle travels approximately 1464.4 meters from touchdown to a complete stop.
d) What is the deceleration 8 seconds after touchdown?
Deceleration is the rate at which velocity decreases. To find the deceleration 8 seconds after touchdown, we can evaluate the derivative of the velocity equation at t=8:
v(t) = 189 - 3t^2
a(t) = dv(t)/dt
Taking the derivative, we get:
a(t) = d(189 - 3t^2)/dt
a(t) = -6t
Plugging in t=8 into the derivative equation, we have:
a(8) = -6(8)
a(8) = -48 meters per second squared
Therefore, the deceleration 8 seconds after touchdown is -48 meters per second squared. Note the negative sign indicates the deceleration.