Also another question:
A particle travelling in a straight line has velocity v m/s at time t s. Its acceleration is given by )
. Its velocity is 50m/s initially and is reduced to 3 m/s. What is the expression for the time taken in seconds for this to occur.
Answer 
i understand everything except the limits of the intergration. why did they use them (50 and 3) to get the time?
Firstly, what is an integral? If you have dy/dx = f'(x), then integrating f'(x) from bounds a to b will give f(b) - f(a), or the change in y.
So similarly here, we have dv/dt = -0.05(v^2-5)
dt/dv = -20/(v^2-5)
Now we want the change in time. Note any similarities with the above? If we integrate the function given from the initial velocity, v=50 to the final velocity, v = 3, we find the time elapsed between those two velocities. So we integrate -20/(v^2-5) dv from 50 to 3.
But VCE doesn't like it when the upper bound is lower, so flip the bounds to get the integral of 20/(v^2-5) dv from 3 to 50.
how did you get that
Let's have two vectors, v and u, and we want to find the "amount" of v that is in the direction of u, so we want to find a constant k such that ku, v and v-ku form a right angled triangle.
If you draw up a triangle, you'll find that this means the vector u and v-ku must be perpendicular.
Therefore u.(v-ku) = u.v - k*u.u = 0
k=u.v/u.u
Vector resolute is hence u.v/u.u * u
Insert vector arrows where appropriate.
yeah but the tank has 20 litres in it already. wouldn't it be 2/20 =1/10?
i think you wouldve been confused by the 10 litres of salt
Inflow rate = 0
Outflow rate? We're removing 2 L of solution per minute while keeping the volume constant. The concentration of salt at any time is x/v, and v is a constant 20, so the concentration is x/20
Now mass flow rate = concentration * volume flow rate= x/20 * 2 = x/10 kilograms / minute
This is the change in mass, or dx/dt, so dx/dt = - x/10
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