A particle moves along a straight line. When it is a distance s from a fixed point, where s > 1, the velocity v is given by v = (3s + 2) / (2s-1). Find the acceleration when s = 2
When I first did this problem I thought that the answer was simply -7/(2s-1)^2 = -7/9= acceleration
But it turns out you have to use acceleration=dv/dt= dv/ds x ds/dt= v x dv/ds
So the correct answer would just be (-7/9) x (3s+2)/(2s-1), substituting 2 for s?
I would appreciate it if you could tell me if the above method ^^^^ is right or if I need to do something else. Thanks!