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?
C) 1metre
d.189-3(8)=165
a.189-3t^2
b. 189-3(1)^2=186t
a) To find the velocity of the shuttle at touchdown, we need to find the derivative of the position function, s(t), with respect to time, t. The derivative of the function s(t) will give us the velocity.
s(t) = 189t - t^3
To find the derivative, we apply the power rule of differentiation. Since the derivative of t^3 is 3t^2 and the derivative of 189t is 189, the derivative of s(t) is:
s'(t) = 189 - 3t^2
Therefore, the velocity of the shuttle at touchdown (t = 0) is:
s'(0) = 189 - 3(0)^2
s'(0) = 189 - 0
s'(0) = 189 meters per second
b) To find the time required for the shuttle to stop completely, we need to find the value of t when the velocity, s'(t), is equal to zero. Setting s'(t) = 0 and solving for t will give us the time required.
s'(t) = 189 - 3t^2
Setting s'(t) = 0:
0 = 189 - 3t^2
Rearranging the equation:
3t^2 = 189
Dividing both sides by 3:
t^2 = 63
Taking the square root of both sides:
t = ±√63
Since time cannot be negative, the time required for the shuttle to stop completely is:
t = √63 ≈ 7.94 seconds
c) To find the distance traveled by the shuttle from touchdown to a complete stop, we need to find the integral of the velocity function over the time interval from 0 to the time it takes for the shuttle to stop, which we found to be approximately 7.94 seconds.
The integral of s'(t) = 189 - 3t^2 will give us the displacement function, s(t). Integrating 189 gives us 189t, and integrating -3t^2 gives us -t^3.
So, the displacement function is:
s(t) = 189t - t^3
To find the distance traveled, we need to evaluate s(t) at the upper limit, which is 7.94 seconds, and subtract the value of s(t) at the lower limit, which is 0 seconds.
Distance traveled = s(7.94) - s(0)
Distance traveled = (189 * 7.94 - (7.94)^3) - (189 * 0 - 0^3)
Distance traveled ≈ 750.29 meters
Therefore, the shuttle travels approximately 750.29 meters from touchdown to a complete stop.
d) To find the deceleration 8 seconds after touchdown, we need to find the second derivative of the position function, s(t), with respect to time, t. The second derivative will give us the acceleration.
The second derivative of the function s(t) is obtained by differentiating the velocity function, s'(t), with respect to time.
s'(t) = 189 - 3t^2
Differentiating s'(t):
s''(t) = -6t
To find the deceleration 8 seconds after touchdown, we substitute t = 8 into the acceleration function:
s''(8) = -6(8)
s''(8) = -48 meters per second squared
Therefore, the deceleration 8 seconds after touchdown is -48 meters per second squared.