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Paha777 [63]
3 years ago
11

A skier starts from rest and uniformly accelerates at 2 m/s2 down a slope. How far has the skier traveled at the end of 20 secon

ds?
Physics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

400 meters

Explanation:

In order to solve this problem, you will want to use an equation to solve for displacement with the ability to calculate with an object at rest. The equation that works best for this is d = 1/2 at².

In the word problem, we were given the following information:

a = 2 m/s²

t = 20 seconds

Now we can plug those values into the equation and solve for d.

d = 1/2 at²

d = 1/2 (2 m/s²)(20 seconds)²

d = 1/2 (2) (400)

d = 1/2 (800)

d = 400 meters

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Answer:

a) v₀ = 9.2 m/s

b) y₀ = 7.9 m

Explanation:

The position of the balls is given by the equation:

y =- \frac{1}{2} gt^2 + v_0 t + y_0

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time t

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a) lets divide (a) in two parts:

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v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}

2. part: At time t from part1 + 1.15s, the first ball should land on the ground.

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This leaves only one unknown: v₀

v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}

b)again, lets divide in two parts

1.part: Where will ball1 be relative to ball2 in 1.15s:

t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2

and how fast will it go:

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2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:

0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2

Now let's use this result in the equation for the first ball:

0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m

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