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3 years ago

Describing Acceleration

Physics

2 answers:

vagabundo [1.1K]3 years ago

6 0

Answer: Negative acceleration occurs when an object slows down in the positive direction.

Positive acceleration occurs when an object speeds up in the positive direction.

Explanation: Acceleration can be positive or negative. A negative acceleration is also known as deceleration. And deceleration occurs when an object is coming to rest. While acceleration simply means that an object is Speeding up.

Negative acceleration occurs when an object slows down in the positive direction.

Positive acceleration occurs when an object speeds up in the positive direction.

Aleonysh [2.5K]3 years ago

5 0

Answer: A, C,D,F

A)Neg. acc. occurs when an object slows down in the pos. direct.

C)Neg. acc. occurs when an object slows down in the neg. direct.

D)Pos. acc. occurs when an object speeds up in the pos. direct.

F) Pos. acc. occurs when an object slows down in the neg. direct.

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5-a). Acceleration is a vector defined as the rate of change of velocity.
Its magnitude has units of [length/time²]. The SI unit is meter/second².
Its direction is the direction in which velocity is increasing.

5-b). The graph says that the object's speed is not changing.
When we look at any time, from zero to almost 50 minutes, the
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4 years ago

A rock is dropped (from rest) off a bridge over the Merrimack River. The falling rock

rewona [7]

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31.25 meters or ~31 meters approximately

Explanation:

Let's see which of the 5 variables we are given since this is a constant acceleration problem.

$v_i \ \ \ \ \ \ t \\ v_f \ \ \ \ \ \triangle x \\ a$

We want to find the height of the bridge, aka the vertical displacement of the rock. Let's set the upwards direction to be positive and the downwards direction to be negative.

We are told that the acceleration is 10 m/s² downward, so we have a = -10 m/s².

We are also told that the time it takes the rock to hit the water is 2.5 seconds. Time is the same regardless of the x- or y- direction, so we can say that t = 2.5 seconds.

Now, we aren't told this directly, but we can figure out that the velocity in the y-direction is 0 m/s, since the rock is dropped from rest off the bridge. Therefore, $v_i=0 \frac{m}{s}$ .

We want to find the vertical displacement, the height of the bridge, so we can say that $\triangle x= \ ?$

We have 4 out of 5 variables:

$v_i,\ a, \ t, \ \triangle x$

Look through the constant acceleration equations to see which equation has all 4 of these variables. You should come up with this one (no final velocity):

$x_f=x_i+v_it+\frac{1}{2}at^2$

Subtract $x_i$ from both sides of the equation to get:

$\triangle x=v_it+\frac{1}{2}at^2$

Substitute in our known variables and solve for delta x.

$\triangle x=(0\frac{m}{s})(2.5s) + \frac{1}{2} (-10\frac{m}{s^2})(2.5s)^2$

0 m/s multiplied by 2.5 s is 0, so we have:

$\triangle x =\frac{1}{2} (-10)(2.5)^2$

Evaluate the exponent first and multiply the terms together.

$\triangle x =(-5)(6.25)$
$\triangle x =-31.25$

The vertical displacement is -31.25 meters from the rock's starting position, so we can say that the height of the bridge is 31.25 meters, which is approximately 31 meters tall.

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