# velocity problem

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.

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.

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. 