I am kinda confused how does negative velocity(-v) and positive acceleration(+a) exactly and vice versa slow down objects? i have a thought about it please let me know if i am correct. according to newtons first law of motion an object at rest or moving at a constant velocity will continue to move at constant velocity unless no external force acts upon it. and force causes acceleration so can i say that imagine an car moving at constant velocity in the west direction due to breakes all of its resultant force acts in the east and so that causes a positive acceleration in east and this stops the car? and am i also correct about the brakes part, and if possible can anyone tell me about the way how brakes work according to physics and which direction is the resultant force in this case of brakes?


2 Answers 2


You seem to be confusing velocity, which is a vector, and its norm.

Newton's first law is about constant vector velocity:

  • constant norm: the object moves with constant speed
  • constant direction: the motion remains along a given direction

Newton's second law states that an object can keep moving with constant velocity either because no force is applied upon it, or because the sum of the forces is zero.

Let's return to your example of a point-like mass moving to the west. Let's define an $x$ axis oriented from east to west. It means that velocity is: $$\vec{v}=v\,\vec{e}_x$$ with $v>0$ (at least at first). For this system to slow down, $v$ has to decrease, which means that the acceleration must be: $$\vec{a}=a\,\vec{e}_x$$ with $a<0$ because acceleration is the derivative of velocity with respect to time, and a decreasing function has a negative derivative.

Newton's second law then states that the object must feel a force with the same direction as the acceleration, which is eastward in the case.

Generally speaking, when you want to find out whether a motion is accelerated or decelerated:

  • If $\vec{a}.\vec{v}>0$, the motion is accelerated.
  • If $\vec{a}.\vec{v}<0$, the motion is decelerated.

This result is probably more natural with kinetic energy, as this energy's derivative yields precisely $\vec{a}.\vec{v}$.


Consider a numerical example.

At time $t=0\,s$ the constant acceleration of a body is $+2\,\rm m\,s^{-2}$ due East whilst it is travelling at a velocity of $6\, \rm m \,s^{-1}$ due West.
So at time $t=0\,s$ the magnitude of the velocity of the body (its speed) is $6\, \rm m \,s^{-1}$.

Using the kinematic equation for constant constant acceleration $v=u+a\,t$ and taking the direction East as positive, after a time of $2\,\rm s$ the velocity of the body is $-6+2 \times 2 = -2 \,\rm m\,s^{-1}$.
So now the speed of the body is $2 \,\rm m\,s^{-1}$ ie the car has slowed down in the sense that its speed has decreased.


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