course Phy 231
This is the first random problem from Assignment 2.
***If the velocity of the object changes from 4 cm / sec to 16 cm / sec in 8 seconds, then at what average rate is the velocity changing?To find the rate of change, subtract the initial velocity from the final velocity and divide by the time period. In this case the initial velocity is 4 cm/s and the final velocity is 16cm/s, and the time period is 8s. So:
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( 16cm/s - 4cm/s ) / 8s = (12cm/s)/8s = 1.5cm/s/s
The average rate of change of velocity is 1.5cm/s/s
***A ball rolling from rest down a constant incline requires 8.2 seconds to roll the 97 centimeter length of the incline. What is its average velocity?
The average velocity is found by the formula Vave = distance/clock time
97cm / 8.2s = about 11.83cm/s
The average veloicty is about 11.83cm/s.
***An object which accelerates uniformly from rest will attain a final velocity which is double its average velocity.
***What therefore is the final velocity of this ball?
If the final velocity is two times the original velocity, then the final velocity is 11.83cm/s * 2
which equals about 23.66cm/s.
***What average rate is the velocity of the ball therefore changing?
Once again the formula for the avg rate of change of velocity is
( Vf - Vi ) / T where Vf = 23.66cm/s, Vi = 0cm/s, and T = 8.2s
Substituting this into the equation tells us that the average rate of change of velocity is about 2.89 cm/s/s.
***An automobile accelerates uniformly down a constant incline, starting from rest. It requires 10 seconds to cover a distance of 132 meters. At what average rate is the velocity of the automobile therefore changing?
To solve this problem we must find the final velocity. To do this, we find the average velocity and multiply it by 2 since it has uniform acceleration.
Vave = 132m / 10s = 13.2 m/s.
Vf = Vave * 2 = 13.2 m/s * 2 = 26.4m/s.
The average rate of change of the velocity then can be found by the formula:
( Vf - Vi ) / T where Vf = 26.4 m/s, Vi = 0 m/s, and T = 10s.
( 26.4 m/s - 0 m/s ) / 10s = 2.64 m/s/s. "
Excellent solutions.
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