How Should a Constant Force Affect an Acceleration-Time Graph?

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A constant force applied to an object results in a constant acceleration, which is represented as a straight, horizontal line above zero on an acceleration-time graph. Since friction is negligible, the acceleration remains steady as long as the force and mass are unchanged. The discussion emphasizes that the actual numerical calculation of acceleration is less important than understanding the relationship defined by Newton's second law (F = ma). The conclusion confirms that the correct representation of the graph is indeed a solid, straight line above the zero mark. This understanding is crucial for accurately depicting motion under constant force conditions.
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Acceleration/time graphs!

Hey all...

I'm having tons of trouble with what seems to be an extremely easy question. :mad:

The question portrays a toy car...which can move in either direction along a horizontal line (the + position axis) The car is given a large force toward the right of constant magnitude is applied to the car. (it says to assume that friction is so small that it can be ignored).

The task is to sketch on an acceleration/time graph a solid line caused by the applied large force.

My guess was that the graph would look like a perfectly straight line somewhere above zero...but I am usually always wrong about this kind of thing :cry:

Can anyone help me? Thanks! :smile:
 
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You have a constant net force, and a constant mass, what do you think the acceleration would be like? Calculate the acceleration of the object for a few different points in time if you need to. The actual calculation won't be important, the important part will be realizing that the actual calculation isn't important. ;)
 
im turning in my homework with the answer to this as a solid, STRAIGHT, and HORIZONTAL line above the zero mark...

just curious - did i get it right? :) ...or wrong? :rolleyes:
 
Right. F = ma you said the force was constant, we know the mass is constant, thus the acceleration must be constant as well.
 

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