Help with deriving time, velocity, displacement

AI Thread Summary
To solve for the time taken by a dog accelerating from 2.0 m/sec to 5.0 m/sec over a distance of 14.0 m, the correct formula is t = d / ((Vi + Vf) / 2). The discussion highlights confusion regarding the algebraic manipulation of the equation, particularly in how to handle the average velocity. Participants emphasize the importance of considering constant acceleration in the calculations. Clarifications are made about the proper use of parentheses in the formula to avoid misinterpretation. Understanding the average velocity is crucial for accurately determining the time taken.
RachaelA
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Homework Statement


A dog accelerates from 2.0 m/sec to 5.0 m/sec while covering a distance of 14.0 m. How long did this take

Homework Equations


d=(Vi+Vf/2)t

The Attempt at a Solution


2(-Vi-Vf)d=t

I am doing something wrong. Please help. Thanks[/B]
 
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Your algebra is off. You shouldn't be subtracting the sum of the velocities. In the equation (Vi+Vf) is being multiplied into time "t." So what should you actually do to move it to the other side?
 
I think I should divide it?
 
RachaelA said:
d=(Vi+Vf/2)t

Is this the formula to find the time ??

You have to put the acceleration in consideration ,

Try to Find the acceleration first [I think it is constant] then the time from the equations of motion with constant acceleration.

:)
 
Last edited:
RachaelA said:

Homework Statement


A dog accelerates from 2.0 m/sec to 5.0 m/sec while covering a distance of 14.0 m. How long did this take

Homework Equations


d=(Vi+Vf/2)t

The Attempt at a Solution


2(-Vi-Vf)d=t

I am doing something wrong. Please help. Thanks[/B]

I don't understand why you used
2(-Vi-Vf)d=t

t is not velocity/distance if there is acceleration.

I am assuming the acceleration to be a constant here. Use the laws of motion to find the acceleration. Once acceleration is determined it is easy to find how long the dog took to reach a final velocity.
 
Sorry. A small correction in my previous reply.
t is not velocity*distance.
 
vktsn0303 said:
Sorry. A small correction in my previous reply.
t is not velocity*distance.
This is the answer according to my teacher

t=d/(Vf+Vi/2)
 
Maged Saeed said:
You have to put the acceleration in consideration ,
Seems like RachaelA has been taught to use the average velocity, being merely the average of initial and final velocities in the case of uniform acceleration. So no need to calculate the acceleration as such.
RachaelA said:
d=(Vi+Vf/2)t
Please use parentheses correctly. What you have written means ##d = \left(V_i+\frac{V_f}{2}\right)t##, but I hope you mean ##d = \left(\frac{V_i+V_f}{2}\right)t##, which you could have written as d=((Vi+Vf)/2)t.
RachaelA said:
I think I should divide it?
Yes.
RachaelA said:
This is the answer according to my teacher
t=d/(Vf+Vi/2)
Again, I hope you mean ##t = \frac d{\left(\frac{V_i+V_f}{2}\right)}## (which can be simplified a little, but no matter).
If so, dividing would have produced that.
 
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