Kinematic body velocity Question

[Arman777]

Homework Statement

A body travels 200 cm, in the first 2 sec,and 220 cm in the next 4 sec What will be the velocity at the end of seventh second from the start ?

Homework Equations

$v_{avg}=\frac {Δx} {Δt}$


3. The attempt at a solutio
$(v_{avg})_1=\frac {Δx} {Δt}$ for first 2 sec is 2m/2s which its 1m/s
$(v_{avg})_2=\frac {Δx} {Δt}$ for next 4 sec which its 2.2m/4s=0.55m/s

So if we assume the body moved from t=6 to t=7 with speed 0.55m/s we get

$(v_{avg})_{tot}=\frac {Δx} {Δt}$

which its Δx=(0.55m/s.5m)+(1m/s.2s)=4.75m
Δt=7s
so the answer is Δx/Δt=4.75m/7s=0.67m/s=67cm/s

the answer says 10cm/s

I don't know where I did go wrong thanks

 

[kuruman]

Here is where you went wrong.

So if we assume the body moved from t=6 to t=7 with speed 0.55m/s

If you assume that from 6 to 7 s the body is moving at constant speed, you are asserting that it is not accelerating, which is clearly incorrect because it is.

 

[Arman777]

Ok,I was wrong

 

[kuruman]

Ok well then what will be the acceleration of the object I found its -0.11 m/s2 but its again doe

Can you show us how you found this acceleration?

 

[Arman777]

The change in velocity is 0.45m/s but what about Δt=? I assumed its 4 sec but that doesn't sound right for me ?

 

[Arman777]

First I thought there must be some acceleraiton ( in the negative way) but that doesn't sound correct for me...If you look carefully the acceleration cannot be constant I guess

 

[kuruman]

You should not be trying to do this using average velocities unless you really, really know what you're doing. Use instead the kinematic equation that gives the position of an object at any time t. You know two pairs of positions and times, so use the math to put the object at the known positions at the known times. Do not assume that the object is at rest at time t = 0 because there is nothing in the problem that says that is.

 

[Arman777]

You should not be trying to do this using average velocities unless you really, really know what you're doing. Use instead the kinematic equation that gives the position of an object at any time t. You know two pairs of positions and times, so use the math to put the object at the known positions at the known times. Do not assume that the object is at rest at time t = 0 because there is nothing in the problem that says that is.

let me try

 

[kuruman]

let me try

OK, but if it doesn't work, show me what you've done so I can troubleshoot it.

 

[Arman777]

I want to present an argument.We know that the object can't turn around, and the question ask us the average velocity.Which its simply Δx/Δt here Δt=7 and the Δx is minimum 4.2 m which 4.2m/7s=0.6 m/s.This means that If its not turning then the average velocity should be higher then 0.6 which it cannot be the answer

For your case,I am not sure do we know the final or initial velocities of the object.(as you said we don't know the objects inital speed so we can't claim anything about it I guess) We just know the distance traveled and how much takes time.
I think from this info we can calculate only average velocity.

 

[Arman777]

If we assume its rest at inital point and the speed of object is 1m/s and the end then from kinematic equations

2m=1m/s.2s+1/2at^2 which a is zero.but again I don't think this is true.I think we have to go from average velocity

 

[Arman777]

If we assume its rest at inital point and the speed of object is 1m/s and the end then from kinematic equations

2m=1m/s.2s+1/2at^2 which a is zero.but again I don't think this is true.I think we have to go from average velocity

or maybe its true...I am so confused for this simple question

 

[Doc Al]

I presume you are to assume constant acceleration?

Use your position as a function of time equations, but do not assume values for the initial speed or acceleration. You have two data points, so you can solve for those values.

 

[gneill]

I want to present an argument.We know that the object can't turn around, and the question ask us the average velocity.

Where does the problem ask for the average velocity? When I read the problem statement I see that it asks for the velocity, which is always taken to be the instantaneous velocity, at a particular point in time.

You are given a pair of data points that each consist of a time and distance. Start by writing the general kinematic equation of motion that gives distance then fill in the given data to form two equations. Yes there are two unknowns: initial velocity and acceleration. But that's why you have two data points to work with so that you can establish two equations and two unknowns.

If you wish to look at the problem in a more geometrical fashion then you should be able to sketch a velocity versus time graph, knowing that the area under the curve gives the distance traveled. You'll end up writing the same equations to solve for the initial velocity and the slope (acceleration) as you would obtain via kinematic equations, but perhaps the visualization of the geometry will help?

 

[Arman777]

Ok,I found it thanks a lot...I don't know why this question confused me a lot...

 

[gneill]

Ok,I found it thanks a lot...I don't know why this question confused me a lot...

It happens sometimes that we follow a train of thought with a fixed idea that is based on an inaccurate assumption. Not a problem if you periodically take a step back to re-evaluate how you got where you ended up!

Just in case anyone is interested in pursuing the geometrical approach to the problem, here's my sketch that presents the problem in that fashion with the given data incorporated. It's just the original problem stated geometrically; the same amount of work is left to do as in the original problem

 

[kuruman]

And just in case anyone is interested in pursuing the solution using average velocities, here is how thanks to gneill's informative graph.

Note that, for constant acceleration, the average velocity is equal to the instantaneous velocity at the midpoint of the time interval.

$(v_{avg})_1=\frac {Δx} {Δt}$ for first 2 sec is 2m/2s which its 1m/s
$(v_{avg})_2=\frac {Δx} {Δt}$ for next 4 sec which its 2.2m/4s=0.55m/s

Thus, the instantaneous velocity is 1 m/s at t = 1 s and 0.55 m/s at t = 4 s. Now one can use $a = \frac{\Delta v}{\Delta t}$ to find the acceleration first. To find the velocity at t = 7 seconds, one may use v = v₀ + a Δt with v₀ = 1 m/s and Δt = 6 s, or v₀ = 0.55 m/s and Δt = 3 s. The answer is the same either way.

This is what I meant when I wrote, "You should not be trying to do this using average velocities unless you really, really know what you're doing."

 

[Arman777]

Thus, the instantaneous velocity is 1 m/s at t = 1 s and 0.55 m/s at t = 4 s. Now one can use a=ΔvΔta=ΔvΔta = \frac{\Delta v}{\Delta t} to find the acceleration first. To find the velocity at t = 7 seconds, one may use v = v0 + a Δt with v0 = 1 m/s and Δt = 6 s, or v0 = 0.55 m/s and Δt = 3 s. The answer is the same either way.

This is what I meant when I wrote, "You should not be trying to do this using average velocities unless you really, really know what you're doing."

I see your idea...Yeah...I don't like it too.Actually my native language is not english.I can understand most problems but in kinematics...I get confused most of the times.My approach was wrong for sure but I was trying to find average velocity during t=0 to t=7