What is the difference between speed and velocity?

Question

So I am having trouble grasping onto the concept of velocity decreasing and how to obtain what time velocity would be decreasing and why. I know these rules but do not understand why this works for speed and not velocity: a(x) X v(x) < 0 means that particle is slowing down a(x) X v(x) > 0 means that particle is speeding up

What does it mean for a velocity function to be decreasing vs the speed function to be decreasing?

Answer 1

For this example, the velocity oscillates between 1 and -1. When the value is positive, the the object is moving along the x axis in the positive direction. Lets call that direction right. When the velocity is negative, the object is moving left. This is because velocity is a vector quantity. This means it has both a magnitude (the speed) and a direction.

The speed, however, is always non-negative. This is because speed is defined as the absolute value of the velocity.

In the example given, for $3 < t < 4.5$ the velocity is decreasing, from $0$ to $-1$. This you should be able to see from the function $v(t) = \sin\left(\frac{\pi}{3}t\right)$ that describes the velocity. However, the speed increases, from the absolute value of $0$ to the absolute value of $-1$; from $0$ to $1$.

Answer 2

Bushra. In th e problem you posted,the object only moves in $x$-direction. In this case it should be easy, the object only moves either to the right or left direction. Here is where velocity and speed can be distinguished.

If the object moves to the right with speed 1m/s, it means the velocity is also 1m/s.

But if the obect move to the left with the same speed, then the velocity is counted as -1m/s.

Velocity is speed with direction.

In your problem, if

$$v(t) = \sin(\pi t/3 )$$ then the speed is $$ |v(t)| = |\sin(\pi t/3)|$$

To determine whether the velocity is decreasing or increasing in time, use first derivative in time :

$$ v'(t)= (\pi/3) \cos (\pi t/3) $$

If $v'(t)>0$ then the velocity is increasing. If $v'(t)<0$ then the velocity is decreasing.

$$ Speed(t) = |v(t)| = |\sin(\pi t/3)| = \begin{cases} \sin(\pi t/3) , \: \: \: \sin(\pi t/3) \ge 0 \\ -\sin(\pi t/3) , \: \: \: \sin(\pi t/3) < 0 \end{cases}$$

$$ Speed'(t) = \begin{cases} (\pi/3) \cos(\pi t/3) , \: \: \: \sin(\pi t/3) \ge 0 \\ -(\pi/3) \cos(\pi t/3) , \: \: \: \sin(\pi t/3) < 0 \end{cases}$$

In interval $[3,4.5]$ we have the interval for $\pi t/3$ is $[\pi, 1.5 \pi]$. At that interval, the $v(t) \le 0$, and $v'(t) \le 0$, so the velocity is decreasing. It keeps moving farther to the left.

Can you conclude for the speed?

---

When an object's speed is decreasing, the object moves slower in time. And the lowest possible speed is 0, which means the object stop moving.

Now for velocity, imagine if the object was initially moving with constant velocity 2m/s, then become to decreasing in time. If it keeps decreasing only until $v=0$, then it has been slowing down until the object stop moving. Now, notice that velocity can also be negative, so if $v$ keeps decreasing pass 0 and becomes negative, this means the object is has been slowing down until $v=0$, but then becomes faster in different direction (to the left). Just like a car slowing down for a U-turn and then speeding up.