Don't know how to calculate the velocity

[brad sue]

Hi,
i don't know how to calculate the velocity.
The problem is:
Use the velocity of a particle as a function of time tabulated below , to calculate the position of the particle at eact of the times ( assume at t=0 ,the particle is at the origin)

Times(s)
0.0
.5
1.5
2.5
3.5

velocity (m/s)
0.00
0.75
1.75
8.75
21.75

position (m)
0.00
0.19
1.44
6.69
21.94


The answer is the column in red, BUT I don't know how to find those values.
please Can someone can help me how to find the values of the positions.

I tried the formula x=v*t but it does not work.

please help

Thank you

Bertrand

 

[TD]

Since the velocity isn't constant, you should think about the acceleration too!

 

[mezarashi]

I tried the formula x=v*t but it does not work.

Yes, that is because this formula that we use in everyday life assumes constant velocity and no acceleration (change in velocity). You can see that in your case the velocity is increasing. Displacement is actually the integral of velocity over time, but a no calculus approach is alright too. You just won't appreciate the meaning of it all as much unfortunately. The best way to find the distance traveled by an object knowing its velocity is to find the area under the curve of the velocity-time graph. Plot it out. Then use basic geometry to find the area of the triangles below the curve.

 

[HallsofIvy]

I'm having trouble understanding this. You say "i don't know how to calculate the velocity." but then "The answer is the column in red, BUT I don't know how to find those values."

The column in red gives distances moved, not velocity!

"I tried the formula x=v*t but it does not work."

Yes, that formula only works for constant velocity which is not the case here.

You can, however, use an "average" velocity between the points where you are given the velocity.

For example, just looking at the first two values:
time velocity
0.0 0.00
.5 0.75

Assuming a constant acceleration (we don't know that's true, but given no other information, that's the simplest thing to do), the "average velocity" is (0.00+ 0.75)/2= 0.375 m/s. At that average speed, for 0.5 s, the object will go (0.375)(0.5)= 0.1875 which, rounded to 2 decimal places, is the 0.19 given!

Now, after another second (t= 1.5 s), the velocity is 1.75 m/s. During that second, the average velocity was (0.75+ 1.75)/2= 1.25 m. At that speed for 1 s, the object will go another (1.25 m/s)(1 s)= 1.25 m. Since it had already gone 0.19 m, that will be 0.19+ 1.25= 1.44! Get the idea?

Between t= 1.5 and t= 2.5, the average velocity is (1.75+ 8.75)/2= 5.25 m. How far with the object go in that second? What is its new position?

 

[brad sue]

thanks

Hi HallsofIvy

OK , to answer your question , the object travels 5.25m ( =5.25*(2.5-1.5))
and the new position is 5.25+1.44=6.69m!

I forgot that d=v*t is fot constant velocity!
(the column in red was the solution from the book!)

Thank you soo much

Brad

I'm having trouble understanding this. You say "i don't know how to calculate the velocity." but then "The answer is the column in red, BUT I don't know how to find those values."

The column in red gives distances moved, not velocity!

"I tried the formula x=v*t but it does not work."

Yes, that formula only works for constant velocity which is not the case here.

You can, however, use an "average" velocity between the points where you are given the velocity.

For example, just looking at the first two values:
time velocity
0.0 0.00
.5 0.75

Assuming a constant acceleration (we don't know that's true, but given no other information, that's the simplest thing to do), the "average velocity" is (0.00+ 0.75)/2= 0.375 m/s. At that average speed, for 0.5 s, the object will go (0.375)(0.5)= 0.1875 which, rounded to 2 decimal places, is the 0.19 given!

Now, after another second (t= 1.5 s), the velocity is 1.75 m/s. During that second, the average velocity was (0.75+ 1.75)/2= 1.25 m. At that speed for 1 s, the object will go another (1.25 m/s)(1 s)= 1.25 m. Since it had already gone 0.19 m, that will be 0.19+ 1.25= 1.44! Get the idea?

Between t= 1.5 and t= 2.5, the average velocity is (1.75+ 8.75)/2= 5.25 m. How far with the object go in that second? What is its new position?