- #1
naggy
- 60
- 0
A particle moves along a straight line with accel. a = 2[tex]\sqrt{v}[/tex] where v = [tex]\frac{dx}{dt}[/tex] is the velocity. At t = 2s, the particle is located at x= (64/3)m with velocity v = 16m/s. Find the location and acceleration at time t = 3s (hint: integrate)
Now, if the differential equation is dv/dt = 2[tex]\sqrt{v}[/tex]. I don´t know how to solve such an equatoin (because of the square root). But then I thought v must be constant for the integration to work, but that would mean that the acceleration would be zero and that makes no sense. I need help.
Now, if the differential equation is dv/dt = 2[tex]\sqrt{v}[/tex]. I don´t know how to solve such an equatoin (because of the square root). But then I thought v must be constant for the integration to work, but that would mean that the acceleration would be zero and that makes no sense. I need help.