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Physics For Scientists & Engineers
Found in: Page 49

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Short Answer

On another planet, a marble is released from rest at the top of a high cliff. It falls 4.00 m in the first 1 s of its motion. Through what additional distance does it fall in the next 1s ? (a) 4.00 m (b) 8.00 m (c) 12.0 m (d) 16.0 m (e) 20.0 m

So, the correct answer is option (c) 12.0 m.

See the step by step solution

Step by Step Solution

Step 1: Equation of freely falling body

Vertical displacement of a freely falling body is given by the relation, x=12agt2 , where Planet ag=acceleration.

Step 2: Explanation for correct answer

In first case, (First 1s)

Travels 4.00 m in the first 1s.

4.00 m=12ag(1 s)2 ag=8.00 m/s2

In second case, (after 2s )

xf=12(8.00 m/s2)(2 s)2 =16 m

Hence, after 2 seconds of travel from rest, the change in position is 16.00 m.

So, the change in position between 1s and 2s is

16.00m-4.00 m=12m.

Hence, the answer is 12 m.

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