Sure. By finding the area under the curve within the limits of time, we can find the distance covered in that time duration. Since v = ds / dt, ds = v dt. So for small time gap dt, vdt gives the displacement ds. So by integrating vdt for the limits, we can get the total displacement S
WFTT-DT was created in 1988.
KPTF-DT was created in 2001-04.
CKSA-DT was created on 1960-09-23.
KGLA-DT was created on 2007-06-05.
1978 to 1981 DT-175 with CDI (no ignition points) spark plug gap .6-.8mm or .024-.032 inch, spark plug# B9ES
1978 to 1981 DT-175 with CDI (no ignition points) spark plug gap .6-.8mm or .024-.032 inch, spark plug# B9ES
1974 yamaha dt 125
I would go with the KTM, but if you do go with the DT, here is a website that helps derestrict the bike for better performance: http://www.dtr125.net/
650 ml3
28
around 5 horse power believe it or not i was very surprised. i only answer this because my ttr 125 4-stroke h my dt 125 2 stroke has 10 hp stock
DT song code send to 53463
Start Here http://www.state.nd.us/ea/teams/dt/st/current.html#csa http://www.ttuhsc.edu/sop/ContinuingEd/hipaa/Gen_GapAnalysis_Remediation.doc
The question is to PROVE that dy/dx = (dy/dt)/(dx/dt). This follows from the chain rule (without getting into any heavy formalism). We know x and y are functions of t. Given an appropriate curve (we can integrate piece-wise if necessary), y can be written as a function of x where x is a function of t, i.e., y = y(x(t)). By the chain rule, we have dy/dt = dy/dx * dx/dt. For points where the derivative of x with respect to t does not vanish, we therefore have (dy/dt)/(dx/dt) = dy/dx.
Rockets work on the conservation of vector energy, cP. 0 = dcP/dr = cdP/cdt=dP/dt = d(mV)/dt = mdV/dt + Vdm/dt=0 Thus, mdV/dt = -Vdm/dt, or (dV/dt)/V = -(dm/dt)/m. The Rocket's mass accelerates at the rate of the mass changes dm/dt.
d/dt cot (t) dt = - cosec2(t)