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KonstantinChe [14]
3 years ago
11

An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt

+1/2at^2
Physics
1 answer:
rusak2 [61]3 years ago
7 0
If the velocity is constant then the acceleration of the object is zero.
a=0 (m/s^2)
Thus when we apply the equation
\Delta X=vt+(at^2/2)
It remains
\Delta X =vt
or equivalent
t=(\Delta X/v) =7500/278 =26.98 (seconds)

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Answer:

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Given that;

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Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

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