# Kinematic body velocity Question

• Arman777
In summary: If you plot velocity versus time and you assume acceleration is constant, then the problem is equivalent to finding the slope of a line that passes through the two given data points. Once you know the slope you can work backwards to find the initial velocity. The slope of the line is the acceleration. You can find the y-intercept of the line and that is the initial velocity. You can find the x-intercept of the line and that is the time when the object was at rest. From that information you can calculate the average speed over the first two seconds. It will equal the distance traveled in the first two seconds divided by the time interval. You can do the same thing for the second data point.
Arman777
Gold Member

## 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

I don't know where I did go wrong thanks

Here is where you went wrong.
Arman777 said:
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.

Thomas Tucker
Ok,I was wrong

Arman777 said:
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?

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 ?

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

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.

Thomas Tucker
kuruman said:
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

Arman777 said:
let me try
OK, but if it doesn't work, show me what you've done so I can troubleshoot it.

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.

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 said:
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

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.

Arman777 said:
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?

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

Arman777 said:
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

Arman777
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.
Arman777 said:
## (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."

gneill
kuruman said:
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

## 1. What is the definition of kinematic body velocity?

Kinematic body velocity is the rate of change of an object's position with respect to time. It is a vector quantity that describes both the speed and direction of an object's motion.

## 2. How is kinematic body velocity different from kinematic body acceleration?

Kinematic body velocity measures the change in an object's position, while kinematic body acceleration measures the change in an object's velocity. Velocity is a first derivative of position, while acceleration is a second derivative.

## 3. What are the units of measurement for kinematic body velocity?

Kinematic body velocity is typically measured in meters per second (m/s) in the metric system, or feet per second (ft/s) in the imperial system.

## 4. How is kinematic body velocity calculated?

Kinematic body velocity can be calculated by dividing the change in an object's position by the change in time. This can also be written as the derivative of position with respect to time.

## 5. Why is kinematic body velocity important in physics and engineering?

Kinematic body velocity is a crucial concept in understanding and predicting the motion of objects. It is used in various fields such as physics, engineering, and robotics to analyze and design systems that involve motion. Understanding kinematic body velocity can also help in optimizing the performance of machines and vehicles.

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