velocity problem

Thread Starter

braddy

Joined Dec 29, 2004
83
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
 

Haus76

Joined Aug 31, 2005
7
Originally posted by braddy@Aug 31 2005, 10:29 AM
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
[post=10052]Quoted post[/post]​


from the tables psted it appears to have an acceleration compontent. X=V*T only works for constant velocity. at this time i cannot recall the correct formula.
 

Raspider

Joined Jan 31, 2007
1
Hello Brady, this problem is rather tricky. The Acceleration is not constant, it is changing therefore your formula x=V*T cannot be correct as it assumes constant acceleration.

Therefore you must use the following formula. By the (0) I mean naught or X2 if you will.

x-x(0) = (V^2 -V(0)^2)/2a)

So for instance assume the particle starts at x=0 V=0 Y=0

At your first time .5s the velocity is .75 M/S. Let's caculate the acceleration which is just the Change in Velocity/ Change in Time. So .75M/S-0M/S / .5M/S-0M/S which gives me 1.5=A

So I plug it into the original formula. (.75M/S)^2 - (0M/S)^ / 1.5(2) which in turn gives me .1875 which they rounded to .19

On the second step you will have to add .19 to the solution you obtain from step 1 to obtain X. A is 1/1 = 1

(1.75M/S)^2-(.75M/S)^2 / 2(A) = 1.25 + .19 = 1.44 which is now X1.

and so on and so forth.
 

recca02

Joined Apr 2, 2007
1,214
i believe the above problem can be also solved with the help of graph
plot graph of vel vs time, the area under the curve gives displ
 

beenthere

Joined Apr 20, 2004
15,819
The acceleration is given by V2 (velocity sub 2) - V1 over T2 - T1, where the velocity sub 2 is the velocity at time 2, and so on.

The velocity at any time is given by the formula - V = V0 (V sub zero) + AT (acceleration times time).
 

recca02

Joined Apr 2, 2007
1,214
it might be very tedious to solve it analytically as
v = u + at
the above formula assumes constant acceleration.
and integration can not be applied since change of acceln as a functn of
time is not known
trying graphically might help
 

noodle

Joined Sep 14, 2007
5
In this case the acceleration is not constant. So break it up into small intervals and apply equation s=ut+1/2at^2, which gives the distance travelled.
Ill illustrate for first two intervals(1) t varies from 0 to 0.5 secs and velocity varies from 0 to 0.75 m/s^2.
Therfore acceleration is change in velocity/time taken a=0.75/0.5=1.5m/s^2
Applying equation s=0+1/2*1.5*0.5^2=0.19m.
(2)t varies from 0.5 to 1.5 secs and velocity varies from 0.75 to 1.75m/s^2
a=1m/s^2.
distance travelled s=0.75*1+1/2*1*1^2=1.25m.
Initially the position of the particle is 0.19m. therfore new position is 1.25+0.19=1.44m. similarly do the rest.:)
 
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