If a ship is moving forward has positive acceleration what can be said about the velocity of the ship?

Physics

1 answer:

ra1l [238]3 years ago

7 0

Increasing

Explanation:

If a ship is moving forward and has a positive acceleration, we can say that the velocity of the ship is increasing.

Velocity is the rate of change of displacement of a body with time.

Acceleration is the rate of change of velocity with time.

Acceleration = $\frac{change in velocity }{time}$

If a ship is moving with positive acceleration, the velocity of the ship is increasing also.

This is because when velocity is increasing, the change will be positive.
From the equation, to have positive velocity, the final velocity must be greater than the initial one.
With this, acceleration will be positive.
So we can say, velocity of the ship is increasing.

learn more:

Acceleration brainly.com/question/3820012

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Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$