Answer 30.8 m s Given: For front car of the train: Initial velocity of the front car as the train start = v f o = 0 m s Final velocity of the front car as it reaches the worker = v f f = 25 m s Initial position of front car as the train starts = x f o = 95 m Final position of front car as it reaches the worker = x f f = 95 + 180 = 275 m Acceleration of the train = a For last car of the train when front car is at the location of the worker: Initial velocity of the last car as front car passes the worker = v l o = 25 m s Final velocity of the last car as it passes the worker = v l f Initial position of last car as front car passes the worker = x l o = 180 m Final position of last car as it reaches the worker = x f f = 95 + 180 = 275 m Acceleration of the train = a Formula Used: Final position during accelerated motion of an object is given as x f f = x f o + v f 2 − v o 2 2 a Calculation: Consider the motion of the front car: Final position of the front car at the position of worker is given as x f f = x f o + v f f 2 − v f o 2 2 a Substituting the values 275 = 95 + 25 2 − 0 2 2 a 275 − 95 = 625 2 a a = 625 2 ( 180 ) a = 1.7 m s 2 Consider the motion of the last car: Final position of the last car at the position of worker is given as x l f = x l o + v l f 2 − v l o 2 2 a Substituting the values 275 = 180 + v l f 2 − 25 2 2 ( 1.7 ) 275 − 180 = v l f 2 − 625 3.4 v l f 2 − 625 = ( 95 ) ( 3.4 ) v l f = 30.8 m s Conclusion: Therefore, speed of last car is 30.8 m s Explanation Given: For front car of the train: Initial velocity of the front car as the train start = v f o = 0 m s Final velocity of the front car as it reaches the worker = v f f = 25 m s Initial position of front car as the train starts = x f o = 95 m Final position of front car as it reaches the worker = x f f = 95 + 180 = 275 m Acceleration of the train = a For last car of the train when front car is at the location of the worker: Initial velocity of the last car as front car passes the worker = v l o = 25 m s Final velocity of the last car as it passes the worker = v l f Initial position of last car as front car passes the worker = x l o = 180 m Final position of last car as it reaches the worker = x f f = 95 + 180 = 275 m Acceleration of the train = a Formula Used: Final position during accelerated motion of an object is given as x f f = x f o + v f 2 − v o 2 2 a Calculation: Consider the motion of the front car: Final position of the front car at the position of worker is given as x f f = x f o + v f f 2 − v f o 2 2 a Substituting the values 275 = 95 + 25 2 − 0 2 2 a 275 − 95 = 625 2 a a = 625 2 ( 180 ) a = 1.7 m s 2 Consider the motion of the last car: Final position of the last car at the position of worker is given as x l f = x l o + v l f 2 − v l o 2 2 a Substituting the values 275 = 180 + v l f 2 − 25 2 2 ( 1.7 ) 275 − 180 = v l f 2 − 625 3.4 v l f 2 − 625 = ( 95 ) ( 3.4 ) v l f = 30.8 m s Conclusion: Therefore, speed of last car is 30.8 m s

6th Edition

ISBN: 9780130606204

Author: Douglas C. Giancoli

Publisher: Prentice Hall

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Question

Chapter 2, Problem 63GP

To determine

The speed of the last car as it passes the worker

Expert Solution & Answer

Answer to Problem 63GP

30.8ms

Given:

For front car of the train:

Initial velocity of the front car as the train start =vfo=0ms

Final velocity of the front car as it reaches the worker =vff=25ms

Initial position of front car as the train starts =xfo=95m

Final position of front car as it reaches the worker =xff=95+180=275m

Acceleration of the train =a

For last car of the train when front car is at the location of the worker:

Initial velocity of the last car as front car passes the worker =vlo=25ms

Final velocity of the last car as it passes the worker =vlf

Initial position of last car as front car passes the worker =xlo=180m

Final position of last car as it reaches the worker =xff=95+180=275m

Acceleration of the train =a

Formula Used:

Final position during accelerated motion of an object is given as

xff=xfo+vf2−vo22a

Calculation:

Consider the motion of the front car:

Final position of the front car at the position of worker is given as

xff=xfo+vff2−vfo22a

Substituting the values

275=95+252−022a275−95=6252aa=6252(180)a=1.7ms2

Consider the motion of the last car:

Final position of the last car at the position of worker is given as

xlf=xlo+vlf2−vlo22a

Substituting the values

275=180+vlf2−2522(1.7)275−180=vlf2−6253.4vlf2−625=(95)(3.4)vlf=30.8ms

Conclusion:

Therefore, speed of last car is 30.8ms

Explanation of Solution

Given:

For front car of the train:

Initial velocity of the front car as the train start =vfo=0ms

Final velocity of the front car as it reaches the worker =vff=25ms

Initial position of front car as the train starts =xfo=95m

Final position of front car as it reaches the worker =xff=95+180=275m

Acceleration of the train =a

For last car of the train when front car is at the location of the worker:

Initial velocity of the last car as front car passes the worker =vlo=25ms

Final velocity of the last car as it passes the worker =vlf

Initial position of last car as front car passes the worker =xlo=180m

Final position of last car as it reaches the worker =xff=95+180=275m

Acceleration of the train =a

Formula Used:

Final position during accelerated motion of an object is given as

xff=xfo+vf2−vo22a

Calculation:

Consider the motion of the front car:

Final position of the front car at the position of worker is given as

xff=xfo+vff2−vfo22a

Substituting the values

275=95+252−022a275−95=6252aa=6252(180)a=1.7ms2

Consider the motion of the last car:

Final position of the last car at the position of worker is given as

xlf=xlo+vlf2−vlo22a

Substituting the values

275=180+vlf2−2522(1.7)275−180=vlf2−6253.4vlf2−625=(95)(3.4)vlf=30.8ms

Conclusion:

Therefore, speed of last car is 30.8ms

Chapter 2, Problem 62GP

Chapter 2, Problem 64GP

Chapter 2 Solutions

Physics: Principles with Applications

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