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Zarrin [17]
3 years ago
12

This is the question with the options

Physics
1 answer:
melamori03 [73]3 years ago
4 0

Answer:

(c) \frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}

Explanation:

Average speed is calculated as (total distance)/(total time). We have three segments in the journey, indexed by 0, 1, and 2:

v = \frac{s}{t}=\frac{v_0t_0+v_1t_1+v_2t_2}{t_0+t_1+t_2}\\t_1=t_2\implies\\v=\frac{v_0t_0+v_1t_1+v_2t_1}{t_0+t_1+t_1}=\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}

We also know that the distance of the first segment is the same as one of segment 2 and 3 together:

v_0t_0=v_1t_1+v_2t_1\\v_0\frac{t_0}{t_1}=v_1+v_2

Going back to the average speed expression, divide by t_1:

\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}=\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}

and combine the two equations:

\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}=\frac{2(v_1+v_2)}{\frac{v_1+v_2}{v_0}+2} = \frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}

The last form matches your choice (c).

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Final kinetic energy before striking the ground = Initial potential energy + Initial kinetic energy

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