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goofy44
28.09.2019 •
Mathematics
Askateboarder starts from rest and moves down a hill with constant acceleration in a straight line for 6 s. in a second trial he starts from rest and moves along the same straight line with the same acceleration for only 2 s. how does his displacement from his starting point compare with the first trial? . a) one-third as large. b) three times larger . c)one-ninth as large. d) nine times larger
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Ответ:
His displacement from his starting point compare with the first trial : C)One-ninth as large.
Further explanationRegular straight motion is the motion of objects on a straight track that has a fixed speed
Formula used
S = distance = m
v = speed = m / s
t = time = seconds
Straight motion changes regularly are the straight motion of objects that have a fixed acceleration
Formula used
V = vo + at
Vt² = vo² + 2as
St = distance on t
vo = initial speed
vt = speed on t
a = acceleration
The skateboarder performs 2 movements without initial speed (vo = 0) so it is considered free fall motion
1. motion decreases with time t = 6sSt1 = vot + 1/2 at² ⇒ vo = 0 so,
St1 = 1/2 .at² ⇒ a = g
St1 = 1/2.g (6²)
St1 = 1/2 g.36
St1 = 18 g
2. motion decreases with time t = 2sSt2= 1/2.g (2²)
St2 = 1/2 g.4
St2 = 2 g
So his displacement from his starting point compare with the first trial :
The correct answer : C)One-ninth as large.
Learn moreThe distance of the elevator
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resultant velocity
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the average velocity
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Keywords: free fall motion , gravitational acceleration, fixed acceleration
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