Help with deriving time, velocity, displacement

In summary: You can verify that by multiplying the right-hand side of what you wrote by ##2##. Doing so would give ##\left(\frac{2d}{V_f+V_i}\right) \cdot \left(\frac{V_f+V_i}{2}\right) = \frac{2d}{2} = d##.
  • #1
RachaelA
3
0

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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  • #2
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?
 
  • #3
I think I should divide it?
 
  • #4
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:
  • #5
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.
 
  • #6
Sorry. A small correction in my previous reply.
t is not velocity*distance.
 
  • #7
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)
 
  • #8
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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FAQ: Help with deriving time, velocity, displacement

How do you derive time, velocity, and displacement?

To derive time, velocity, and displacement, you need to use the basic kinematic equations, which involve the variables of initial velocity (u), final velocity (v), acceleration (a), time (t), and displacement (s). These equations are:

1) v = u + at

2) s = ut + 1/2at^2

3) v^2 = u^2 + 2as

4) s = 1/2(u+v)t

By rearranging these equations and plugging in known values, you can solve for the desired variable.

What units are used for time, velocity, and displacement?

Time is typically measured in seconds (s), velocity in meters per second (m/s), and displacement in meters (m). However, depending on the specific situation, these units may vary. For example, time may be measured in minutes (min) or hours (hrs) for longer durations, and velocity may be measured in kilometers per hour (km/h) for faster speeds.

Can you derive time, velocity, and displacement without knowing the acceleration?

Yes, it is possible to derive time, velocity, and displacement without knowing the acceleration. However, you would need to have at least two of the other variables (initial velocity, final velocity, or displacement) in order to use the kinematic equations and solve for the unknown variable. If you only have one known variable, it is not possible to accurately derive the others.

How is time, velocity, and displacement related?

The relationship between time, velocity, and displacement is described by the basic kinematic equations. These equations show how changes in one variable (such as acceleration) can affect the others. For example, if acceleration increases, velocity and displacement will also increase over time. Similarly, if acceleration is constant, velocity will change at a constant rate, and displacement will change at an increasing rate.

Can you derive time, velocity, and displacement for non-uniform motion?

Yes, the kinematic equations can be used to derive time, velocity, and displacement for non-uniform (changing) motion. However, since the equations assume constant acceleration, they may not provide completely accurate results. In these cases, more complex mathematical models may be needed to accurately calculate these variables.

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