How to calculate the average velocity?

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Indranil
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Homework Statement


The position of an object moving along the x-axis is given by x=a+bt^2, where a=8.5m,b=2.5 ms^-2, and t is measured in seconds. what is the average velocity between t=2.0 s and t= 4.0 s

Homework Equations


as we know, Vavg = xf-xi / tf-ti (f= final position and time and i= initial position and time)

The Attempt at a Solution


Where to put the values because we have the values of t but there are no values of x
 
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Indranil said:
Where to put the values because we have the values of t but there are no values of x
Your problem statement has an equation for x(t) does it not?
 
gneill said:
Your problem statement has an equation for x(t) doe it not?
No, the only equation is 'x=a+bt^2'
 
average velocity between t=2.0 s and t= 4.0 s
So find x when t=2 and find x when t=4, then you'll have all the data you need.
 
Merlin3189 said:
So find x when t=2 and find x when t=4, then you'll have all the data you need.
I am doing this below
x=a +bt^2 =8.5 +2.5 x 4 =18.5 (when t=4)
x=at + bt^2 =8.5 + 2.5 x 2 =13.5 (when t=2)
see the values below
(a=8.5m, b=2.5 ms^-2, t=2.0 s and t= 4.0 s)
Now what to do?
 
Merlin3189 said:
1- what is average velocity?

2- what do your two x values tell you?
1. Average velocity = TotalDisplacement / TotalTime
2. The initial position and the final position
If I do the math
Vavg=18.5-13.5 /4-2 = 5/2 (but the answer does not match the answer in my book. in my book, my answer is 15m/s)
Now, what to do?
 
Indranil said:
x=a +bt^2 =8.5 +2.5 x 4 =18.5 (when t=4)
x=at + bt^2 =8.5 + 2.5 x 2 =13.5 (when t=2)
see the values below
(a=8.5m, b=2.5 ms^-2, t=2.0 s and t= 4.0 s)
Sorry, I didn't check this before. You forgot to square the time.

Also there is some confusion between, x=a +bt^2 and x=at + bt^2

In the question statement, you gave the first version.

But your last post is correct in your statement of average velocity and working from your previous results. You just need to check those first calculations.
 
Merlin3189 said:
Sorry, I didn't check this before. You forgot to square the time.

Also there is some confusion between, x=a +bt^2 and x=at + bt^2

In the question statement, you gave the first version.

But your last post is correct in your statement of average velocity and working from your previous results. You just need to check those first calculations.
It would be x=a+bt^2
 
Ok. That's what it said in the Q and it is consistent with the units given for a and b.

So now you can get the right values if you stick to this equation and take care with squaring t.
 
Merlin3189 said:
Ok. That's what it said in the Q and it is consistent with the units given for a and b.

So now you can get the right values if you stick to this equation and take care with squaring t.
I did but still, the answer does not match the answer in my book.
As you said I have done below
X=18.5 when t=2 and X=40 when t==4 now according to the average velocity
Vavg=xf-xi / tf-ti = 21.5 /2 ms^-1
 
Indranil said:
I did but still, the answer does not match the answer in my book.
As you said I have done below
X=18.5 when t=2 and X=40 when t==4 now according to the average velocity
Vavg=xf-xi / tf-ti = 21.5 /2 ms^-1
Double check your position calculation for t = 4 s.
 
gneill said:
Double check your position calculation for t = 4 s.
Thank you for finding my mistake. I checked it and it's done. Now the answer is 15m/s. Thank you all for your kind efforts.