What the motion of an object that has an acceleration of 0 m/s

Physics

1 answer:

Degger [83]3 years ago

8 0

<span>Everything in the system is stable and therefore the objects motion is stable. That is to say it is not changing what it is already doing. As far as i know zero times zero is still zero. In that case then the motion must be constant or stable.</span>

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The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 12t + 6 for t ≥ 0

avanturin [10]

Answer:

(a) $v(t) = 6t^2 - 6t - 12$ , $a(t) = 12t - 6$

(b) When $0 \leq t < 0.5$ , object is slowing down, when $t > 0.5$ object is speeding up.

Explanation:

(a) To get the velocity function, we need to take the derivative of the position function.

$v(t) = \frac{ds(t)}{dt} = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12$

To get the acceleration function, we need to take the derivative of the velocity function.

$a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6$

(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0

$12t - 6 < 0 \\12t < 6 \\t < 0.5$

On the other hand, object is speeding up when a > 0

$12t - 6 > 0 \\12t > 6 \\t > 0.5$

Therefore, when $0 \leq t < 0.5$ , object is slowing down, when $t > 0.5$ object is speeding up.

6 0

3 years ago

Knowing the constant g what will the gravitational force between two masses be if the gravitational force between them is 36n an

charle [14.2K]

The gravitational force between two masses is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the two masses
r is the separation between the two masses

We see that the force is proportional to the inverse of the square of the distance: $F \sim \frac{1}{r^2}$
therefore, if the distance is tripled:
r'=3r
The force decreases by a factor 1/9:
$F \sim \frac{1}{(3r)^2}= \frac{1}{9} \frac{1}{r^2}$

Since the original force was 36 N, the new force will be
$F' = \frac{1}{9} (36 N)= 4 N$

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3 years ago

Each blade of a fan has a radius of 11 inches. If the fan’s rate of turn is 1440o /sec, find the following. (a) The angular spee

notka56 [123]

Answer:

a) 24.43 radians per second

b) 268.73 inches per second

Explanation:

a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

$\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}$