Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
D. The flow of energy,heat, and work
