Initial Velocity and Acceleration: A Curious Conundrum

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How is V(initial) = 0 when Acc at start is equal to 4m/s^2

shouldnt there be some type of Velocity?
 
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  • Textbook page, Example 3 Minimum Length for a Runway: left column solution showing a_x = 4.00 m/s^2, v_x0 = 0 and x_0 = 0 boxed in red, right column equations v_x = a_x t and x = 1/2 a_x t^2 with results 20.0 m/s and 50.0 m at t = 5.00 s, and 613 m runway length.
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Nope.

Lets say you drop a ball from your hand. The acceleration is 9.8m/s^s downward, but when the ball is in your hand it is not moving. (unless of course you are swinging your arm).

So at initial, the very beginning, there is no velocity. Velocity changes due to acceleration.
 
I thought this was only true for "free falling" objects...

For example... I posted some graphs...
graph 1 is velocity increasing... then graph 2 is acceleration based off graph 1...


However if velocity had a constant slope say y = 3.. then acc would be zero...
My thoughts where... if initial acceleration is not zero in word problem then initial velocity must not be zero... which i showed in my two graphs...
Obviously I am wrong.. I just need to wrap my mind around it ..
 
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  • Two side-by-side axis graphs labelled 1 and 2. Graph 1 shows a straight line rising from the origin with positive slope. Graph 2 shows a horizontal line at constant height.
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Acceleration is a change in velocity during some time interval. In order for any object to begin moving from rest, whatever the direction of travel, the object must undergo acceleration.

When you are in a car stopped at a traffic light, and the car takes off after the light changes, the force you feel pushing you into the seat back is due to the car accelerating away from the traffic light.