Help me understand the equation V = u + delt v

  • Thread starter Indranil
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  • #1
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Homework Statement


V = u + delta v
[Here V = final velocity, u = initial velocity and delta v = change in velocity (v-u/t-t)]
As I know, a = v-u/t so, v = u + at
So could you explain why v = u + delta v?

Homework Equations


V = u + delta v
[Here V = final velocity, u = initial velocity and delta v = change in velocity (v-u/t-t)]
As I know, a = v-u/t so, v = u + at
So could you explain why v = u + delta v?

The Attempt at a Solution


As I know, a = v-u/t so, v = u + at
So could you explain why v = u + delta v?
 

Answers and Replies

  • #2
TSny
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So could you explain why v = u + delta v?

delta v (or Δv) is defined to be the change in velocity. Change in velocity, in turn, is defined to be the final velocity (V) minus the initial velocity (u). So, by definition,

Δv = V - u.

Rearrange this to arrive at what you want to show.

Note that this result does not depend on using any acceleration equations. In particular, the acceleration need not be constant. The expression for Δv is valid for any motion along a straight line. The expression is a direct consequence of the meaning of "change in velocity".

For motion in 2 or 3 dimensions of space, there is a similar expression except that you would be working with velocity vectors:

##\Delta \vec v = \vec V - \vec u##

This is often written

##\Delta \vec v = \vec v_f - \vec v_i## where ##\vec v_i## and ##\vec v_f## are the initial and final velocity vectors, respectively. On the right hand side, you would be doing vector subtraction. This equation can be rearranged to get

##\vec v_f =\vec v_i + \Delta \vec v##
 
  • #3
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4,992

Homework Statement


V = u + delta v
[Here V = final velocity, u = initial velocity and delta v = change in velocity (v-u/t-t)]
As I know, a = v-u/t so, v = u + at
So could you explain why v = u + delta v?

Homework Equations


V = u + delta v
[Here V = final velocity, u = initial velocity and delta v = change in velocity (v-u/t-t)]
As I know, a = v-u/t so, v = u + at
So could you explain why v = u + delta v?

The Attempt at a Solution


As I know, a = v-u/t so, v = u + at
So could you explain why v = u + delta v?
Would it work better for you if we wrote ##\Delta v=at##?
 
  • #5
177
11
delta v (or Δv) is defined to be the change in velocity. Change in velocity, in turn, is defined to be the final velocity (V) minus the initial velocity (u). So, by definition,

Δv = V - u.

Rearrange this to arrive at what you want to show.

Note that this result does not depend on using any acceleration equations. In particular, the acceleration need not be constant. The expression for Δv is valid for any motion along a straight line. The expression is a direct consequence of the meaning of "change in velocity".

For motion in 2 or 3 dimensions of space, there is a similar expression except that you would be working with velocity vectors:

##\Delta \vec v = \vec V - \vec u##

This is often written

##\Delta \vec v = \vec v_f - \vec v_i## where ##\vec v_i## and ##\vec v_f## are the initial and final velocity vectors, respectively. On the right hand side, you would be doing vector subtraction. This equation can be rearranged to get

##\vec v_f =\vec v_i + \Delta \vec v##
You said above 'the acceleration need not be constant' so there is no any point using the acceleration, but what about the time (t)? as we know, delta v = vf-ui / t. Could you explain this part, please?
 
  • #6
mjc123
Science Advisor
Homework Helper
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As has been said above, Δv = v - u, i.e. the change in velocity. That's what the symbol Δ means. It is NOT (v-u)/t, which is the average acceleration. If it was, you couldn't say v = u + Δv because the units of u and Δv wouldn't match.
 
  • #7
177
11
delta v (or Δv) is defined to be the change in velocity. Change in velocity, in turn, is defined to be the final velocity (V) minus the initial velocity (u). So, by definition,

Δv = V - u.

Rearrange this to arrive at what you want to show.

Note that this result does not depend on using any acceleration equations. In particular, the acceleration need not be constant. The expression for Δv is valid for any motion along a straight line. The expression is a direct consequence of the meaning of "change in velocity".

For motion in 2 or 3 dimensions of space, there is a similar expression except that you would be working with velocity vectors:

##\Delta \vec v = \vec V - \vec u##

This is often written

##\Delta \vec v = \vec v_f - \vec v_i## where ##\vec v_i## and ##\vec v_f## are the initial and final velocity vectors, respectively. On the right hand side, you would be doing vector subtraction. This equation can be rearranged to get

##\vec v_f =\vec v_i + \Delta \vec v##
I don't understand this part 'Note that this result does not depend on using any acceleration equations. In particular, the acceleration need not be constant.' Could you please get this point easier for me so that I could understand the concept better?
 
  • #8
CWatters
Science Advisor
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Gold Member
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If my initial velocity (u) is 10mph and I change my velocity (delta velocity) by +20mph my final velocity (v) is 30mph.

u + delta v = v
10 + 20 = 30

It's that simple.

The equation says nothing about how quickly the velocity changes (aka acceleration).

There are other equations that do relate initial and final velocity and time to acceleration.
 

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