The question is incomplete. Here is the complete question:
A lunar lander is descending toward the moon’s surface. Until the lander reaches the surface, its height above the surface of the moon is given by y(t)=b−ct+dt2, where b = 800 m is the initial height of the lander above the surface, c = 60.0 m/s, and d=1.05m/s2 .
(a) What is the initial velocity of the lander, at t = 0?
(b) What is the velocity of the lander just before it reaches the lunar surface?
Answer:
(a) - 60 m/s
(b) - 15.48 m/s
Explanation:
part (a)
In order to find initial velocity at time t = 0 we need to differentiate the expression for height with respect to time. The expression of height is given in the question.
Velocity = dh / dt
velocity = d (b−ct+dt²) / dt
after differentiating with respect to t we get
velocity = - c + 2dt
substituting the value of c, d and t.
velocity = - (60) + 2 (1.05) (0)
velocity = - 60 m/s (Answer)
part (b)
We can find the answer of velocity by first finding the time taken to reach the lunar surface.
Just before it reaches the lunar surface means that its height above the surface of moon is about to become 0. We can find time of the journey by equating the expression of height above the moon with 0.
y(t) = b−ct+dt²
y(t) = 0
b-ct+dt² = 0
rearranging the equation
dt² – ct + b = 0
This is a quadratic equation. We can use quadratic formula to solve the equation for time.
Using the equation
t = (c ± √(c² – 4 db)) / 2d
where,
d = 1.05
c = 60.0
b = 800
The answers come out to be
t = 21.2 sec or t = 35.9 sec
By using the expression of velocity found above (part ‘a’) we can find the value of velocity
velocity = - c + 2dt
velocity = - 60 + 2 (1.05) (21.2)
velocity = - 15.48 m/s (Answer)
OR
v = - 60 + 2 (1.05)(35.9)
v = 15.39 m/s
We neglect second answer v = 15.39 m/s because we know that velocity cannot be positive in this case as the lunar lander is moving towards moon away from earth.
