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.

Answer 1

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