-3
$\begingroup$

If a force is applied on a body, it will accelerate the body. Due to this acceleration the body will be in motion. So what will be the velocity of that body? Will it be constant?

$\endgroup$

closed as unclear what you're asking by John Rennie, ja72, Jon Custer, Kyle Kanos, heather Jan 31 '17 at 15:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ The definition of acceleration is non-constant velocity in magnitude and/or direction. $\endgroup$ – ja72 Jan 30 '17 at 16:35
  • 1
    $\begingroup$ When exactly do you mean? Of course while it accelerates, the velocity isn't constant. Do you mean afterwards? $\endgroup$ – Steeven Jan 30 '17 at 19:10
  • $\begingroup$ This question requires a bit more detail. What reference frame will you use to make the measurements of position, velocity, and acceleration? Are you in the same reference frame from which the force is applied? $\endgroup$ – David White Jan 31 '17 at 1:49
1
$\begingroup$

Since $\vec{F}=m\times\vec{a}$ ($\vec{F}$ being the force, $m$ the mass of the body, and $\vec{a}$ the resulting acceleration on the body), if a constant force is applied on a body you will have a constant acceleration. A constant acceleration means that you will not have a constant velocity ($v$), but instead a linearly increasing one as time ($t$) passes:

$a=\dfrac{dv}{dt}=\dfrac{F}{m}=const. \rightarrow dv=const.\times dt \rightarrow v(t)=const.\times t $

Of course this is valid only if the magnitude of the velocity is not near the speed of light ($v<<c$), otherwise we should take account of the relativistic effects.

$\endgroup$
0
$\begingroup$

Correct me if I am wrong: I think he meant wether or not the body stops. The answer is no because Newtons first law of motion requires an equal and opposite force to stop the object in motion.

$\endgroup$
  • $\begingroup$ 1) I think the question is formulated on a way that the OP thinks more in an aristotelan mechanics as in the Newtonian. Thus, this question doesn't have very clear answer, first his misunderstandings have to be cleared. 2) Newton's first law isn't so strong imho. $\endgroup$ – peterh Jan 31 '17 at 2:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.