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3 years ago

Kayla starts at -3, walks 5 blocks right and 3 blocks left. What is her displacement?

Physics

1 answer:

Elanso [62]3 years ago

7 0

Answer: The displacement is 1 block.

Explanation:

Let's define:

The right is the positive side.

The left is the negative side.

Then if you start at position A, and you walk N blocks to the right, the new position is:

A + N

And if you start at position A, and you walk M blocks to the left, the new position is:

A - M.

In this case, we know that Kayla starts at -3 and she walks 5 blocks to the right.

Then her new position is:

-3 + 5 = 2

Now she walks 3 blocks to the left, then her new position is:

2 - 3 = -1

The displacement will be equal to the difference between the final position (-1) and the initial position (-2)

Then the displacement is:

D = -1 - (-2) = -1 +2 = 1

The displacement is 1 block.

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Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
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The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

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The centre of mass of the system is the point at which the total

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Here the term Σ mₐ xₐ is called the first moment of the system and the

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Therefore, from this theory we can say that the moment of the system

remain unchanged while calculating the COM.

Hence, The centre of mass of the system is the point at which the total

mass of system could be concentrated without changing the moment

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