A car travels at 15 m/s for 30 minutes. How far did it travel?
1 answer:
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Answer:
2.84 g's with the remaining 1 g coming from gravity (3.84 g's)
Explanation:
period of oscillation while waiting (T1) = 2.45 s
period of oscillation at liftoff (T2) = 1.25 s
period of a pendulum (T) =2π.
where
- L = length
- a = acceleration
therefore the ration of the periods while on ground and at take off will be
=(2π
) / (2π
)
where
- a1 = acceleration on ground while waiting
- a2 = acceleration during liftoff
=
squaring both sides we have
=
=
assuming that the acceleration on ground a1 = 9.8 m/s^{2}
=
a2 = 9.8 x
substituting the values of T1 and T2 into the above we have
a2 = 9.8 x
a2 = 9.8 x 3.84
take note that 1 g = 9.8 m/s^{2} therefore the above becomes
a2 = 3.84 g's
Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.