A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2, and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?
Solution:
We are given the following values: a=−2.10×104m/s2;t=1.85×10−3s;vf=0m/s.
The formula in solving for the initial velocity is
Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of g (9.80 m/s2) ?
Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his
(a) acceleration and
(b) deceleration.
Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.
Solution:
Part A
The formula for acceleration is
aa=ΔtΔv=tf−t0vf−v0
Substituting the given values
aa=5.00sec282m/s−0m/s=56.4m/s2(Answer)
Part B
The deceleration is
aa=1.40s0m/s−282m/s=−201.43m/s2(Answer)
In expressing the computed values in terms of g, we just divide them by 9.80.
The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model, you can view hydrogen, the simplest atom, as having a single electron in a circular orbit 1.06×10-10 m in diameter.
(a) If the average speed of the electron in this orbit is known to be 2.20×106 m/s, calculate the number of revolutions per second it makes about the nucleus.
(b) What is the electron’s average velocity?
Solution:
Part A
The formula to be used is
average speedr=timedistance=td
Rearranging the formula–solving for the distance
d=r×t
Substituting the given values for 1 second period
d=(2.20×106m/s)(1s)=2.20×106meters
This is the total distance traveled in 1 sec.
With the given radius, the total distance traveled in 1 revolution is
1revolution=2πr=πd=π(1.06×10−10m)
Therefore, the total number of revolutions traveled in 1 second is
no. of revolutions=distance in 1 revolutiontotal distance=π(1.06×10−10)2.20×106=6.61×1015revolutions(Answer)
Part B
In one complete revolution, the electron will go back to its original position. Thus, there is no net displacement. Therefore,
A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity
(a) for each of the three intervals and
(b) for the entire motion.
Solution:
Part A
The average velocity for each interval is computed using the formula
v=ΔtΔx
For the first interval
v1=2.50sec15.0meters=6.00m/s(Answer)
For the second interval
v2=1.75sec−3.00meters=−1.71m/s(Answer)
For the third interval
v3=5.20sec21.0m=4.04m/s(Answer)
Part B
For the entire motion, we need displacement from the beginning to the end.
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 3.00×108 m/s.
Solution:
The total distance traveled is computed using the formula
distance=speed×time
Therefore, the total distance traveled is
distance=(3.00×108m/s)(2.56sec)=768000000meters
Hence, the distance between the Earth and Moon is equal to the total distance traveled divided by 2.
The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?
Solution:
The time of travel is computed based on the formula
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