Question

a skateboarder 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

Answer 1

When a skateboarder starts from rest and moves down a hill with constant acceleration in a straight line for 6 seconds in a second trial he starts from rest and moves along the same straight line with the same acceleration for only 2 seconds. His displacement from his starting point compared with the first trial is one-ninth as large. This will make the answer choice C. 

Answer 2

His displacement from his starting point compare with the first trial : C)One-ninth as large.

Further explanation

Regular straight motion is the motion of objects on a straight track that has a fixed speed

Formula used

$\large{\boxed{\bold{S=v\times\:t}}}$

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

$\large{\boxed{\bold{St=vo.t+\frac{1}{2}at^2}}}$

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 = 6s

St1 = 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 = 2s

St2= 1/2.g (2²)

St2 = 1/2 g.4

St2 = 2 g

So his displacement from his starting point compare with the first trial :

$\frac{St_2}{St_1} =\frac{2g}{18g}$

$\frac{St_2}{St_1}=\frac{1}{9}$

The correct answer : C)One-ninth as large.

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Keywords: free fall motion , gravitational acceleration, fixed acceleration