How Is Distance Calculated Using Integration in Nail Making Machine Mechanics?

[fable121]

The velocity of a automatic chuck for moving the wire through a nail making machine may be found from the expression:
v=0.1t(1-t) where v is velocity (m/s) and t is time (sec)

i) using integration calculate the distance moved by the chuck between two times when velocity is 0.

what I did first was change it to 0.1t - 0.1t²

and when t = 1 you get (0.1*1²/2), which = 0.05 and when t = 0 obviously the answer will be 0
So I've got the distance moved as 0.05m

Not really sure if that's right, any help would be great :)

ii) Verify the results obtained in part (i) by using the approximate numerical method known as the Trapezoidal Rule.

Not quite as sure with this one

http://[URL=http://imageshack.us][PLAIN]http://img244.imageshack.us/img244/5751/calky2.png [/URL][/PLAIN] I think this has something to do with it, I'm also pretty sure I'm meant to use the values 0, 0.2, 0.4, 0.6, 0.8 and 1 with the formula v=0.1t(1-t) those values being t

As I said I'm a bit unsure so any help would be great :)

 

[nrqed]

The velocity of a automatic chuck for moving the wire through a nail making machine may be found from the expression:
v=0.1t(1-t) where v is velocity (m/s) and t is time (sec)

i) using integration calculate the distance moved by the chuck between two times when velocity is 0.

what I did first was change it to 0.1t - 0.1t²

and when t = 1 you get (0.1*1²/2),

How did you get this?? To find the distance moved, you have to integrate \int v dt between the initial and final times.

 

[saket1991]

I think he's right...He has put v=dx/dt [After finding values of t from v=0.1t(1-t)...he got t=0,1]
NOw step by step...
dx/dt=0.1t(1-t)
=> dx= [0.1t(1-t)]dt
INtegrating it You get,
[x] (limits are 0,x) = 0.1[(t^2 - t^3 /2)] (limits are 0,1)---------------(1)

In equation (1) its (t raise to power 3)Divided by 2

Solve it and u get the answer.(x=0.05m)

 

[Dick]

I think he's right...He has put v=dx/dt [After finding values of t from v=0.1t(1-t)...he got t=0,1]
NOw step by step...
dx/dt=0.1t(1-t)
=> dx= [0.1t(1-t)]dt
INtegrating it You get,
[x] (limits are 0,x) = 0.1[(t^2 - t^3 /2)] (limits are 0,1)---------------(1)

In equation (1) its (t raise to power 3)Divided by 2

Solve it and u get the answer.(x=0.05m)

Integrating t does not give t^2 nor does t^2 give t^3/2. What are you thinking of?