Context:

In my textbook it is given: _{'momentum' short for 'linear momentum'}:

Mass = $m$, momentum is $p=mv$. In time $\Delta t$, momentum changes by $\Delta p$, the rate of change of momentum is:

 

$$\frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

My Doubts:

1. Isn't a $\Delta$ sign missing beside the $t$ in the second fraction, and thus it should be $\frac{\Delta(mv)}{\Delta t}$
2. How did they derive the third fraction from the second. I tried a lot but can't seem to get that.

My Work:

I have looked at this question - How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?, but it's a totally different equation.

My Final Question:

Can someone please clear my doubts about this equation and help me understand how does:

$$\frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

Thanks a lot !

Context:

In my textbook it is given: _{'momentum' short for 'linear momentum'}:

Mass = $m$, momentum is $p=mv$. In time $\Delta t$, momentum changes by $\Delta p$, the rate of change of momentum is:

$$\frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

My Doubts:

1. Isn't a $\Delta$ sign missing beside the $t$ in the second fraction, and thus it should be $\frac{\Delta(mv)}{\Delta t}$
2. How did they derive the third fraction from the second. I tried a lot but can't seem to get that.

My Work:

I have looked at this question - How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?, but it's a totally different equation.

My Final Question:

Can someone please clear my doubts about this equation and help me understand how does:

$$\frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

Thanks a lot !

Context:

In my textbook it is given: _{'momentum' short for 'linear momentum'}:

Mass = $m$, momentum is $p=mv$. In time $\Delta t$, momentum changes by $\Delta p$, the rate of change of momentum is:

$$\frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

My Doubts:

1. Isn't a $\Delta$ sign missing beside the $t$ in the second fraction, and thus it should be $\frac{\Delta(mv)}{\Delta t}$
2. How did they derive the third fraction from the second. I tried a lot but can't seem to get that.

My Work:

I have looked at this question - How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?, but it's a totally different equation.

My Final Question:

Can someone please clear my doubts about this equation and help me understand how does:

$$\frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

Thanks a lot !

Context:

In my textbook it is given: _{'momentum' short for 'linear momentum'}:

Mass = $m$, momentum is $p=mv$. In time $\Delta t$, momentum changes by $\Delta p$, the rate of change of momentum is:

$$\frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

My Doubts:

1. Isn't a $\Delta$ sign missing beside the $t$ in the second fraction, and thus it should be $\frac{\Delta(mv)}{\Delta t}$
2. How did they derive the third fraction from the second. I tried a lot but can't seem to get that.

My Work:

I have looked at this question - How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?, but it's a totally different equation.

My Final Question:

Can someone please clear my doubts about this equation and help me understand how does:

$$\frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$

Thanks a lot !