How Do You Calculate Acceleration from a Velocity-Time Graph?

[rashad764]

Homework Statement

find the accleration at t=3 seconds

Homework Equations

vf-vi/tf-ti

The Attempt at a Solution

velocity at 3sec is 2 m/s and at 0 sec is at 4m.s
time is 3-0
2-4/3-0, [/B]

 

[Ray Vickson]

1. The problem statement, all variables and g iven/known data
View attachment 210424
find the acceleration at t=3 seconds

Homework Equations

vf-vi/tf-ti

The Attempt at a Solution

velocity at 3sec is 2 m/s and at 0 sec is at 4m.s
time is 3-0
2-4/3-0, [/B]

Your answer of $2 - \frac{4}{3} - 0 = 0.66667$ is incorrect. Even if you put in parentheses (which you should always do) you would get $(2-4)/(3-0) = -0.66667$, which is still incorrect. Can you see where your error lies?

 

[rashad764]

Your answer of $2 - \frac{4}{3} - 0 = 0.66667$ is incorrect. Even if you put in parentheses (which you should always do) you would get $(2-4)/(3-0) = -0.66667$, which is still incorrect. Can you see where your error lies?

is it suppose to be -2?

 

[Ray Vickson]

is it suppose to be -2?

You tell me.

 

[rashad764]

You tell me.

well since the line is going downwards, it has to be negative velocity

 

[CWatters]

No the _velocity_ is still positive at t=3 sec. If you meant negative acceleration you would be correct.

 

[rashad764]

i still can't see what's wrong
velocity at 3seconds is 2 and at 0 seconds its 4

time is 3 seconds minus 0

 

[CWatters]

The problem is that in post #3 you just asked if the answer was -2m/s without showing your working. That's what Ray meant by you tell me.

 

[CWatters]

velocity at 3seconds is 2 and at 0 seconds its 4

time is 3 seconds minus 0

The acceleration at t=3 is the slope of the line at t=3. Does the slope start at t=0?

or look at it this way...

The slope/acceleration is the same at t=2.5 and 3.5sec right. Does your method give the same answer?

 

[rashad764]

The acceleration at t=3 is the slope of the line at t=3. Does the slope start at t=0?

or look at it this way...

The slope/acceleration is the same at t=2.5 and 3.5sec right. Does your method give the same answer?

The problem is that in post #3 you just asked if the answer was -2m/s without showing your working. That's what Ray meant by you tell me.

i was saying if the the velocity was -2 so it would be -2-4/3

acceleration begins at t-2 and and ends at t=4
so it would be( 2-4)/(3-2)

 

[CWatters]

The velocity is never -2m/s. The velocity is positive at all times.

The _change_ in velocity is -2. Is that what you mean?

Sorry for editing this reply. (2-4)/(3-2) = -2 is correct.