How Do You Calculate Acceleration from Velocity?

[Shay10825]

find the acceleration. PLEASE HELP!

Hi

Find the acceleration.

from x= 2m to x= 8m during a 2.5-s time interval.

I found the slope to be 2.8 so the velocity is 2.8. Now how do I fond the acceleration? I know you have to find the slope of the velocity to get the acceleration, but i keep geeting the wrong answer .

The answer is .32 m/s^2

~Thanks

 

[Tide]

I think we're missing some information or you provided incorrect numbers.

If the acceleration is 0.32 m/s^2 then an object will move 1 m in 2.5 seconds under uniform acceleration and starting from rest.

 

[Shay10825]

the problem said:

A particle confined to motion along the x-axis moves with constant acceleration from x= 2m to x= 8m during a 2.5-s time interval. The velocity of the particle at x=8 m is 2.8 m/s. What is the acceleration during this time interval?

I know how to do it with the kinematic equations but I want to do it without them. How would you do it with derivatives?

 

[HallsofIvy]

You left the "The velocity of the particle at x=8 m is 2.8 m/s." out of the original post- that's the "missing information" Tide was talking about.

If we take the constant acceleration to be "a", and initial velocity to be "v₀", then the accleration after t seconds is at+ v₀ and the postion is
(1/2)at²+ v₀t+ 2 (the original position was x= 2).

You are told that when t= 2.5, (1/2)a(2.5²)+ v₀(2.5)+ 2= 8 and a(2.5)+ v₀= 2.8.

Solve those two equations for a and v₀.
at+ v_0[/b

 

[Shay10825]

Sorry .
How come when you find the slope of the v= 2.8 and t=2.5 it does not equal the acceleration of .32 ?

 

[needhelpperson]

because you need another point to determine the slope, not just when v = 2.8. and t = 2.5. You can determine the initial velocity and use that as the second point.

 

[Shay10825]

so (Vf-Vi)/2.5 = acceleration?

but how come (2-8)/2.5 does not = 2.8? How would I fond the slope for the velocity?

 

[needhelpperson]

so (Vf-Vi)/2.5 = acceleration?

but how come (2-8)/2.5 does not = 2.8? How would I fond the slope for the velocity?

because there's an acceleration, the slope for velocity is always changing, thus, you can find the average velocity using (y2-y1)/(x2-x1).

-->(8-2)/(2.5-0) = average velocity.

 

[Shay10825]

How would I find the slope or derivative of the velocity at 2.8m/s and t=2.5 (the answer is .32m/s^2 but how would we find this ).

 

[needhelpperson]

How would I find the slope or derivative of the velocity at 2.8m/s and t=2.5because it is .32m/s^2.

no need for calculus here. Besides, they already told you the slope for the velocity at t = 2.5. It's 2.8m/s. But that's not what you're asking for, since .32m/s^2 is the acceleration.

figure out vi using (vi+vf)/2 = average velocity.

You already know the average velocity.

once you have it. a = (vf-v1)/(2.5)

 

[Shay10825]

I understand all of that but say if this was a different problem and they did not give you the 2.8. How would you find it (2.8) using derivitaves and the information (2 m, 8 m and 2.5 sec)?

 

[needhelpperson]

I understand all of that but say if this was a different problem and they did not give you the 2.8. How would you find it (2.8) using derivitaves and the information (2 m, 8 m and 2.5 sec)?

Then you wouldn't be able to solve it. There's not enough information given.

 

[Shay10825]

so if you had A particle confined to motion along the x-axis moves with constant acceleration from x= 2m to x= 8m during a 2.5-s time interval. Find the velocity of the particle at x=8 m you would not be able to solve it because there is not enough info?

 

[needhelpperson]

so if you had A particle confined to motion along the x-axis moves with constant acceleration from x= 2m to x= 8m during a 2.5-s time interval. Find the velocity of the particle at x=8 m you would not be able to solve it because there is not enough info?

that's correct.

 

[Shay10825]

But I thought the slope or derivative of the displacement or position time graph is the velocity at that time? If it did not say "a 2.5 s time interval" but said specific points then would you be able to find the derivative or the velocity?

 

[needhelpperson]

But I thought the slope or derivative of the displacement or position time graph is the velocity at that time? If it did not say "a 2.5 s time interval" but said specific points then would you be able to find the derivative or the velocity?

Yes, then can you figure it out... if you're given a graph and all...

 

[Shay10825]

THANK YOU SOOOOOO MUCH! I just started AP Physics B and it's all new and so hard .