A passenger, after traveling 35 m while standing on the walkway, starts to walk at a speed of 0.60 m/s relative to the surface of the walkway.How long does she take to travel the total distance of the walkway? Help. Thanks.
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To answer this question start with the basic equation Distance = Velocity * Time (D = V*t). D and V are given and you need to solve for t. So t = D/V . For the first section of travel D = 35 m and V = 0.3 m/s. This gives:
t1= D1/V1 = 35 m / 0.3 m/s = 116.7 s
For the second section of travel V2 = speed of walkway + speed of person walking = 0.3 m/s + 0.6 m/s = 0.9 m/s. D2 is the distance left on the walkway which is 25 m (60 m - 35 m). This gives:
t2 = D2/V2 = 25 m / 0.9 m/s = 27.8 s
The total travel time will be t1 + t2 = 116.7s + 27.8s = 114.4s
This is assuming that the person is walking in the same direction of travel as the walkway (note this is not stated).
I hope this helps.
t1= D1/V1 = 35 m / 0.3 m/s = 116.7 s
For the second section of travel V2 = speed of walkway + speed of person walking = 0.3 m/s + 0.6 m/s = 0.9 m/s. D2 is the distance left on the walkway which is 25 m (60 m - 35 m). This gives:
t2 = D2/V2 = 25 m / 0.9 m/s = 27.8 s
The total travel time will be t1 + t2 = 116.7s + 27.8s = 114.4s
This is assuming that the person is walking in the same direction of travel as the walkway (note this is not stated).
I hope this helps.
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18s just standing
35m is .41666% of the walkway and is walking at 54 m/s during that time
7.499 m/s is how long she was standing there before walking
so i think it took her 8.147s because at 54 m/s it only took her .61s to move across 35m
35m is .41666% of the walkway and is walking at 54 m/s during that time
7.499 m/s is how long she was standing there before walking
so i think it took her 8.147s because at 54 m/s it only took her .61s to move across 35m
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35/.3+25/.9