0
$\begingroup$

I am confused. How does negative velocity and positive acceleration and vice versa slow down objects? Here are my thoughts. Please let me know if I am correct.

According to Newton's first law of motion, an object at rest or moving at a constant velocity will continue to move at constant velocity unless an external force acts upon it. And force causes acceleration. So can I say that if a car moving at constant velocity westwards brakes, all of its resultant force acts in the east and so that causes a positive acceleration in east and this stops the car?

Am I also correct about the brakes part? And if possible, can anyone tell me which direction the resultant force is in this case of brakes?

$\endgroup$
2
  • $\begingroup$ Voting to reopen. A perfectly clear question with several answers below. $\endgroup$
    – gandalf61
    Commented May 3 at 6:27
  • $\begingroup$ @gandalf61 Agreed. I have tidied up the question. $\endgroup$ Commented May 3 at 9:19

3 Answers 3

1
$\begingroup$

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 $\vec{a}.\vec{v}$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

To apply your rule of thumb "negative velocity and positive acceleration and vice versa slow down objects" you must adopt a consistent positive direction for both velocity and acceleration. In you example - braking car moving west - if you choose west as the positive direction then you have positive velocity and negative acceleration. But you could choose east as the positive direction, in which case you have negative velocity and positive acceleration. In either case, the car slows down.

To apply this rule of thumb then velocity and acceleration must be in parallel directions. There are, of course, many situations where velocity and acceleration are not parallel e.g. a car going round a corner.

Brakes work by slowing the rate at which the car's wheels are turning. This does not slow the car directly, but it does create create friction between the car's tyres and the road. This friction is the external force that actually slows the car down. If there is little or no friction between the car's tyres and the road (e.g. if the road is very wet or icy or the tyres are too smooth) then the brakes will not be as effective at slowing the car.

$\endgroup$

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