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?
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
a) To find the velocity of the shuttle at touchdown, we need to find the derivative of the function s(t) with respect to t.
s(t) = 189t - t^3
Taking the derivative of this function, we get:
s'(t) = 189 - 3t^2
To find the velocity at touchdown, we can substitute t = 0 into this derivative since touchdown occurs at t = 0.
s'(0) = 189 - 3(0)^2
s'(0) = 189 - 0
s'(0) = 189
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.
Setting s'(t) = 0 and solving for t:
0 = 189 - 3t^2
3t^2 = 189
t^2 = 189/3
t^2 = 63
Taking the square root of both sides:
t = ±√63
t ≈ ±7.937
Since time cannot be negative, the time required for the shuttle to stop completely is approximately 7.937 seconds.
c) To find the distance traveled from touchdown to a complete stop, we need to integrate the velocity function over the interval from t = 0 to t = 7.937.
∫[0 to 7.937] s'(t) dt = ∫[0 to 7.937] (189 - 3t^2) dt
Integrating the function:
∫[0 to 7.937] (189 - 3t^2) dt = [189t - t^3/3] evaluated from 0 to 7.937
(189 * 7.937 - (7.937)^3/3) - (189 * 0 - 0^3/3)
(1499.893 - 66.778) - (0 - 0)
1433.115 - 0
The shuttle travels approximately 1433.115 meters from touchdown to a complete stop.
d) To find the deceleration 8 seconds after touchdown, we can find the derivative of the velocity function s'(t) with respect to t.
s'(t) = 189 - 3t^2
Taking the derivative of this function, we get:
s''(t) = -6t
To find the deceleration 8 seconds after touchdown, we substitute t = 8 into this derivative.
s''(8) = -6(8)
s''(8) = -48
Therefore, the deceleration 8 seconds after touchdown is -48 m/s^2.
To answer these questions, we need to differentiate the given function s(t) to find the velocity v(t) and acceleration a(t) of the shuttle at any given time t.
a) The velocity of the shuttle at touchdown can be obtained by finding v(0), which is the derivative of s(t) with respect to t evaluated at t = 0.
v(t) = d/dt [189t - t^3] = 189 - 3t^2
v(0) = 189 - 3(0)^2 = 189 - 0 = 189 meters per second
Therefore, the velocity of the shuttle at touchdown is 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 v(t) reaches zero.
v(t) = 189 - 3t^2 = 0
Solving this equation for t, we get:
3t^2 = 189
t^2 = 63
t = sqrt(63)
The time required for the shuttle to stop completely is approximately t ≈ 7.937 seconds.
c) To find the distance traveled from touchdown to a complete stop, we need to integrate the velocity function v(t) with respect to t from t = 0 to t = sqrt(63).
∫v(t) dt = ∫(189 - 3t^2) dt
Integrating, we get:
s(t) = 189t - t^3 + C
To find the value of C, we substitute t = 0 in the above equation:
s(0) = 189(0) - (0)^3 + C
C = 0
Therefore, the equation for s(t) becomes:
s(t) = 189t - t^3
Now, we find the distance traveled from touchdown to a complete stop:
∫[0 to sqrt(63)] (189t - t^3) dt
Evaluating this integral, we get:
s(sqrt(63)) - s(0) ~= 135.81 meters
Therefore, the shuttle travels approximately 135.81 meters from touchdown to a complete stop.
d) To find the deceleration 8 seconds after touchdown, we need to differentiate the velocity function v(t) to obtain the acceleration a(t).
a(t) = d/dt [189 - 3t^2] = -6t
At 8 seconds after touchdown, t = 8. We substitute this value into the acceleration function:
a(8) = -6(8)
a(8) = -48 meters per second squared
Therefore, the deceleration 8 seconds after touchdown is -48 meters per second squared.