The answer is D-Testable
Hope this helps
Till the time car is just adjacent to the bicycle we can say
distance moved by cycle = distance moved by car
Time taken by car to accelerate from rest
![t = \frac{v_f - v_i}{a}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Ba%7D)
![t = \frac{49 - 0}{7} = 7 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B49%20-%200%7D%7B7%7D%20%3D%207%20s)
Time taken by cycle to accelerate
![t = \frac{23 - 0}{15} = 1.53 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B23%20-%200%7D%7B15%7D%20%3D%201.53%20s)
now the distance moved by cycle in time "t"
![d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B23%20%2B%200%7D%7B2%7D%2A1.53%20%2B%2023%28t%20-%201.53%29)
distance moved by car in same time
![d = \frac{7t + 0}{2}(t)](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B7t%20%2B%200%7D%7B2%7D%28t%29)
now make them equal
![3.5t^2 = 17.595 - 35.19 + 23t](https://tex.z-dn.net/?f=3.5t%5E2%20%3D%2017.595%20-%2035.19%20%2B%2023t)
![3.5 t^2 - 23t + 17.595 = 0](https://tex.z-dn.net/?f=3.5%20t%5E2%20-%2023t%20%2B%2017.595%20%3D%200)
![t = 5.68 s](https://tex.z-dn.net/?f=t%20%3D%205.68%20s)
so cycle will move ahead of car for t = 5.68 s
Answer:..
1.One light-year = 63,240 AU
2. It’s about 93 million miles (150 million km),or
one astronomical unit (AU) = 92,955,807 miles (149,597,871 km).
A classic puzzle...
She either kicked it at a wall <em>exactly</em><em /> 10 foot in front of her, where the ball rebounded off the wall.
Or, she kicked the ball straight up, vertically, at a <em>90 degree angle,</em> where due to the law of gravity, which states that anything that goes up must come down, when the soccer ball reaches exactly 10 feet, it falls back down.
(Note: This is nearly impossible to achieve -- exactly 10 feet.)
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