1. CALC A lunar lander is descending toward the moon’s surface. Until the lander
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Question
1. CALC A lunar lander is descending toward the moon’s surface. Until the lander reached the surface, its height above the surface of the moon is given by y(t) = b - ct + dt^2, where b = 800 m is the initial height of the lander above the surface, c = 60.0 m/s and d = 1.053 m/s^2. (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?
2. A rocket carrying a satellite is accelerating straight up from the earth’s surface.At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight (b) the first 5.90-s of its flight.
3.
The initial velocity of train moving along a straight line is 10 m/s and its retardation is 2 m/s^2 when it approaches the station. Calculate the distance moved by the train in the fifth second of its motion.
Explanation / Answer
1)Given,
y(t) = b - ct + dt^2
b = 800 m ; c = 60 m/s ; d = 1.053 m/s^2
(a)We know that velocity is:
v = dy/dt = 0 - c + 2dt
for t = 0 ; v = -c = -60 m/s
Hence, at t = 0, v = -60 m/s
b)We need to know the time when it reaches the lunar surface. For this, setting y(t) = 0
dt^2 - ct + b = 0
1.053 t^2 - 60 t + 800 = 0
the above quadratic eqn gives us:
t = 35.7 and 21.3 s
So t = 21.3 s at which it reaches the surface
v = 0 - 62 + 2 x 1.053 x 21.3 = -17.14 m/s
Hence, v = -17.14 m/s
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