Problem 23PE Problem (a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s2 . How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?
Read moreTable of Contents
Textbook Solutions for College Physics
Question
Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light \(\left(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\).
Solution
Step 1 of 2
We need to calculate the distance between the earth and the moon when an astronaut hears an echo after 2.56 s. An echo is the reflection of sound back to the source. Thus, to hear an echo, the sound has to travel from the source to the surface of reflection and then back to the source.
Therefore let us assume the following after neglecting the time delays, atmospheric perturbations, orbital perturbations etc.
t=2T
t is the time taken for the echo to be heard.
T is the time taken for radio waves to travel the one side distance.
d=2D
d is the distance the radio waves travel to generate an echo.
D is the distance the radio waves travel from source to earth.
Given data:
d=?
t=2.56 s
\(s=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) is the speed of radio waves.
full solution