A snail is climbing up a square window of height $20.0 c m$ , starting at the lower left corner and climbing towards the upper right corner.
If the magnitude of the snail's speed is $0.300 c m / s$ , how long will it take for the snail to traverse the window?

133 s

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94.3 s

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77.0 s

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66.7 s

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47.1 s

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Solution

The correct option is D $133 s$
Total distance covered by the snail $d = 20.0 + 20.0 = 40.0$ cm
Constant peed of the snail $v = 0.300$ cm/s
Thus time taken by the snail to reach the upper right corner $t = \frac{d}{t} = \frac{40.0}{0.300} = 133$ s

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A snail is climbing up a square window of height 20.0cm, starting at the lower left corner and climbing towards the upper right corner.If the magnitude of the snail's speed is 0.300cm/s, how long will it take for the snail to traverse the window?

The correct option is D 133sTotal distance covered by the snail d=20.0+20.0=40.0 cmConstant peed of the snail v=0.300 cm/sThus time taken by the snail to reach the upper right corner t=dt=40.00.300=133 s

A snail is climbing up a square window of height $20.0 c m$ , starting at the lower left corner and climbing towards the upper right corner.
If the magnitude of the snail's speed is $0.300 c m / s$ , how long will it take for the snail to traverse the window?

The correct option is D $133 s$
Total distance covered by the snail $d = 20.0 + 20.0 = 40.0$ cm
Constant peed of the snail $v = 0.300$ cm/s
Thus time taken by the snail to reach the upper right corner $t = \frac{d}{t} = \frac{40.0}{0.300} = 133$ s