velocity and acceleration involving integration (1 Viewer)

[AFGHAN22]

1. i) a particle is moving along the x axis. At time t, it is distant x from O and moving with velocity v= dx/dt.
show that the acceleration dv/dt = v dv/dx.
ii) the accelration of a body moving in a straight line is proportional to the velocity, i.e. dv/dt=kv, where k is a constant. when x=0, v=7 and when x=3, v=13. find the value of v when x=2. (hint, use the result in i) )

 

[Riviet]

For part i), just substitute v= dx/dt into v dv/dx, in which the dx's cancel out, giving you dv/dt. You could set it out by starting with RHS and work your way to LHS.

 

[insert-username]

1. i) a particle is moving along the x axis. At time t, it is distant x from O and moving with velocity v= dx/dt.
show that the acceleration dv/dt = v dv/dx.

Let v = dx/dt

Therefore a = dv/dt

Using the chain rule: dv/dt = dv/dx . dx/dt

But dx/dt = v

Therefore acceleration = v.dv/dx


I_F

 

[Mountain.Dew]

for the 1st part, its simply a matter of:

dv/dt = dv / dx * dx / dt = v dv / dx. done and done.

no need to go through all the 'LHS/RHS' stuff, or the

" Let v = dx/dt, Therefore a = dv/dt, etc..." i think its assumed in the question already.

now, for 2nd part, dv/dt=kv

using part 1, kv = v dv / dx so dv/dx = k ==> v = kx + c, x=0, v=7, x=3, v=13. substitute those values in to get simultaenous equations, solve to get c and k, then sub x=2 to get v!