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3 years ago

_________ acceleration occurs when an object speeds up.

Physics

2 answers:

Angelina_Jolie [31]3 years ago

7 0

Positive acceleration occurs when an object speeds up.

Explanation:

Acceleration is the rate of change in velocity per unit time:

$a=\frac{\Delta v}{\Delta t}$

where $\Delta v$ is the change in velocity while $\Delta t$ is the time interval. From the formula, we can understand that when $\Delta v$ is positive, so when the object is speeding up, the acceleration is positive; on the contrary, when $\Delta v$ is negative (so, when the object is slowing down), the acceleration is negative.

fiasKO [112]3 years ago

7 0

The Positive acceleration of a body is the acceleration that speeds up the object under acceleration.

Explanation:

The acceleration of a body is a vector quantity that requires the direction along with the magnitude for its representation. The acceleration of a body is defined as the change in velocity per unit time of the body under the motion.

The direction of acceleration helps in determining whether the body will speed up or its speed will decrease during the motion under the acceleration.

If the direction of acceleration of the body is in the direction opposite to the motion of the body, then this acceleration of body will reduced the velocity of the object.

On the other hand, if the direction of acceleration of the body is same as the direction in which the body is moving, then it will speed up the object during the motion.

The acceleration of the body acting in the direction opposite to the motion of the body is called as the negative acceleration whereas the acceleration of the body along the direction of motion of the body is termed as the positive acceleration.

Thus, The Positive acceleration of a body is the acceleration that speeds up the object under acceleration.

Learn More:

1. Minimum coefficient of the static friction between feet and floor brainly.com/question/1621778

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3. Stress in the bungee cord: brainly.com/question/12985068

Answer Details:

Grade: High School

Subject: Physics

Chapter: Acceleration

Keywords:

acceleration, speed, velocity, positive, speeds up, direction, opposite, along, motion, reduce, increase.

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a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

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The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

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Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

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For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

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