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January 20th, 2017, 06:59 AM   #1
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Could someone give me a hint on how to get the initial velocity

I'm working on some homework. I need to find the initial velocity. The question I'm working on is stated below. After some google research and Khan academy I thought that this was easy so I substituted 0 for time t and came to the results of -50m. I then tried to solve for acceleration and that came out to be zero. These solution just don't feel right to me like I did something wrong, it my gut feeling. The (initial velocity iV) should be distance divided by time right? so V is also zero despite a distance of negative 50 being covered......

Please guide me.

int (Time) t = 0;
int (distance) m = -50;
int (acceleration) a = 0;


Quote:
An experiment done on the Moon shows the position of a particle moving along the y-axis
varies in time according to the expression Y(t)= -50 + 100t ā€“ 20t^2, where Y is in m and t is in
seconds.
(a) Determine the initial velocity, initial position and acceleration of the particle?

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January 20th, 2017, 07:30 AM   #2
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Quote:
An experiment done on the Moon shows the position of a particle moving along the y-axis varies in time according to the expression Y(t)= -50 + 100t ā€“ 20t^2, where Y is in m and t is in seconds.
(a) Determine the initial velocity, initial position and acceleration of the particle?
$y(t) = y_0 + v_0 \cdot t - \dfrac{1}{2}gt^2$

$y_0$ is the initial position relative to some designated zero position

$v_0$ is the initial velocity

$g$ is the magnitude of the acceleration due to gravity

I have a problem with the coefficient of the quadratic term ... $-20 = -\dfrac{1}{2}g \implies g = 40$.
On the Moon, the value of $g$ is known to be about $1.6 \, m/s^2 = 5.3 \, ft/s^2$
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January 20th, 2017, 07:46 AM   #3
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I assume it was intended that y(t) = -50 + 100t - 20t^2.

It follows that y'(t) = 100 - 40t and y''(t) = -40.

Hence y'(0) = 100, y(0) = -50, and y''(0) = -40.

It just remains to add the appropriate units.
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January 20th, 2017, 07:59 AM   #4
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Quote:
Originally Posted by skeeter View Post
$y(t) = y_0 + v_0 \cdot t - \dfrac{1}{2}gt^2$

$y_0$ is the initial position relative to some designated zero position

$v_0$ is the initial velocity

$g$ is the magnitude of the acceleration due to gravity

I have a problem with the coefficient of the quadratic term ... $-20 = -\dfrac{1}{2}g \implies g = 40$.
On the Moon, the value of $g$ is known to be about $1.6 \, m/s^2 = 5.3 \, ft/s^2$
My initial velocity is 100 m/s.

How were you able to pick apart the equation to tell what was the initial velocity?
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January 20th, 2017, 08:08 AM   #5
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Quote:
Originally Posted by The_Ys_Guy View Post
My initial velocity is 100 m/s.

How were you able to pick apart the equation to tell what was the initial velocity?
Usually physics students are exposed to the basic kinematics equation

$x(t) = x_0 + v_0 \cdot t + \dfrac{1}{2}at^2$

modified for motion under the influence of gravity ...

$y(t) = y_0 + v_0 \cdot t - \dfrac{1}{2}gt^2$

... so all you have to do is recognize the values of the respective coefficients and what they represent.


If you are familiar with the relationship between position, velocity, and acceleration derived from the calculus, then have a look at post #3 for that explanation.
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