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ziro4ka [17]
4 years ago
5

An object moves from the position +16m to the position +43m in 12s. What us the total displacement

Physics
1 answer:
NeX [460]4 years ago
3 0

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

                                     =43m -16m

                                     =27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V =  (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12  ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

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Explanation:

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x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=54.5m/s is the projectile's initial speed

\theta=35\° is the angle

t=2.80s is the time since the projectile is launched until it strikes the target

x  is the final horizontal position of the projectile (the value we want to find)

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the projectile (we are told it  was launched at ground level)

y  is the final height of the projectile (the value we want to find)

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x=125m   (4)  This is the horizontal final position of the projectile

For y (2):

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stiks02 [169]

Answer:

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Explanation:

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Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

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Answer:

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