Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
2
Describing Motion: Kinematics in One Dimension
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2-1 to 2-3: Speed and Velocity
2-4: Acceleration
2-5 and 2-6: Motion at Constant Acceleration
2-7: Falling Objects [neglect air resistance]
2-8: Graphical Analysis

Problem 15
A
6.73m/s6.73 m/s
Giancoli 6th Edition, Chapter 2, Problem 15 solution video poster
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VIDEO TRANSCRIPT

When this ball rolls towards the pin it travels the full length of as lane which sixteen and a half meters, but we don’t know the speed that it goes and we don’t know how long it takes to do that. On the way back the sound is making the return trip and it goes the same distance, the length of the alley, back to the ears of the bowler and it travels at a speed of three hundred and forty meters per second but we don’t know how long it takes either. We do know the total time for the ball to leave the bowler, hit the pins, and then for the sound to return. That total is two and a half seconds. So total time ‘tT’ is: ‘t1’ plus ‘t2’, which is two and a half seconds, we call this equation 1. So we can make a system of equations and then solve it for our unknown which is ‘v1’. We have ‘d1’ equals: ‘v1’ times ‘t1’, we call this equation 2. Equation 3 is: ‘d2’ equals ‘v2’ times ‘t2’. Writing these with as many numbers as possible, equation 1 is: ‘t1’ plus ‘t2’ is two and a half seconds. Equation 2 is: sixteen and a half meters equals ‘v1’ times ‘t1’, we don’t know either of those variables. Equation 3 is: ‘d2’ which is sixteen and a half meters equals three hundred and forty meters per second times ‘t2’. Let’s solve equation three, so we’ll call it equation 3b for ‘t2’: ‘t2’ is sixteen and a half meters divided by three hundred and forty meters per second which is zero point zero four eight five two nine seconds. Then equation 1b will solve equation 1, we’ll solve for ‘t1’: ‘t1’ is two and a half seconds minus ‘t2’ which is two and a half minus ‘t2’, zero point zero four eight five two nine seconds giving us ‘t1’ as two point four five one five seconds. And then equation 2b which solves equation 2 which has our actual target, we’ll solve for it: ‘v1’ is sixteen and a half meters divided by ‘t1’ which is sixteen point five meters divided by two point four five one five seconds giving us six point seven three meters per second. That’s the speed of the bowling ball on the way to the pin.

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