Answer
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Hint:-In all the cases passengers have to move the fixed distance , hence distance formula in terms of the velocity and time is used . If only escalator moves it will be the speed of escalator ,if only man moves it will be the speed of man & if man walks up the moving escalator then speed of both is added , and then use the distance formula in terms of the velocity and time and find the value of time.
Complete step-by-step solution:
Distance (\[s\]) covered in all the cases is the same & the distance formula in terms of the velocity and time is given as \[s = vt\].
Case (1) a passenger standing on an escalator to reach the top in 1 minute.
1 minute = 60 sec.
Using the distance formula in terms of the velocity and time.
\[ \Rightarrow s = {v_1}{t_1}\]
\[ \Rightarrow {t_1} = 1\min = 60\sec \]
Speed of the escalator is
\[ \Rightarrow {v_1} = \dfrac{s}{{60}}\]
Case (2) if the escalator does not move, it takes him 3 minutes to walk up.
\[ \Rightarrow {t_2} = 3\min = 180\sec \]
Usi9ng distance formula in terms of the velocity and time
\[ \Rightarrow s = {v_2}{t_2}\]
Speed of the man is
\[ \Rightarrow {v_2} = \dfrac{s}{{180}}\]
Case (3) How long will it take for the passenger to arrive at the top if he walks up the moving escalator.
\[ \Rightarrow V = {v_1} + {v_2}\]
\[ \Rightarrow V = \dfrac{s}{{60}} + \dfrac{s}{{180}}\]
Net speed of man on moving with running escalator is
\[ \Rightarrow V = \dfrac{{3s + s}}{{180}} = \dfrac{{4s}}{{180}}\]
Using the distance formula in terms of the velocity and time
\[ \Rightarrow s = Vt\]
\[ \Rightarrow s = \dfrac{{4s}}{{180}}t\]
Equating it with time
\[ \Rightarrow t = \dfrac{{180s}}{{4s}}\]
\[ \Rightarrow t = 45\sec \]
Hence ,it will take 45sec for the passenger to arrive at the top if he walks up the moving escalator.
Note:- The speed , velocity acceleration all are relative terms , it means they depend on frame of reference.For example : if a man is driving the car with his family on a road trip ,then no matter how much fast the car is going all the members of family are in rest , but when the car crosses the another car moving in just opposite direction of the road it will seem like the car is going faster than an observer on road will see the another car.
Complete step-by-step solution:
Distance (\[s\]) covered in all the cases is the same & the distance formula in terms of the velocity and time is given as \[s = vt\].
Case (1) a passenger standing on an escalator to reach the top in 1 minute.
1 minute = 60 sec.
Using the distance formula in terms of the velocity and time.
\[ \Rightarrow s = {v_1}{t_1}\]
\[ \Rightarrow {t_1} = 1\min = 60\sec \]
Speed of the escalator is
\[ \Rightarrow {v_1} = \dfrac{s}{{60}}\]
Case (2) if the escalator does not move, it takes him 3 minutes to walk up.
\[ \Rightarrow {t_2} = 3\min = 180\sec \]
Usi9ng distance formula in terms of the velocity and time
\[ \Rightarrow s = {v_2}{t_2}\]
Speed of the man is
\[ \Rightarrow {v_2} = \dfrac{s}{{180}}\]
Case (3) How long will it take for the passenger to arrive at the top if he walks up the moving escalator.
\[ \Rightarrow V = {v_1} + {v_2}\]
\[ \Rightarrow V = \dfrac{s}{{60}} + \dfrac{s}{{180}}\]
Net speed of man on moving with running escalator is
\[ \Rightarrow V = \dfrac{{3s + s}}{{180}} = \dfrac{{4s}}{{180}}\]
Using the distance formula in terms of the velocity and time
\[ \Rightarrow s = Vt\]
\[ \Rightarrow s = \dfrac{{4s}}{{180}}t\]
Equating it with time
\[ \Rightarrow t = \dfrac{{180s}}{{4s}}\]
\[ \Rightarrow t = 45\sec \]
Hence ,it will take 45sec for the passenger to arrive at the top if he walks up the moving escalator.
Note:- The speed , velocity acceleration all are relative terms , it means they depend on frame of reference.For example : if a man is driving the car with his family on a road trip ,then no matter how much fast the car is going all the members of family are in rest , but when the car crosses the another car moving in just opposite direction of the road it will seem like the car is going faster than an observer on road will see the another car.
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