Saved Bookmarks
| 1. |
A particle moving along x-axis has acceleration f, at time t, given by f= f_(0) (1 - (t)/(T)), where f_(0) and T are constants. The particle at t= 0 has zero velocity. In the time interval between t= 0 and the instant when f= 0, the particle's velocity (v_(x)) is |
|
Answer» `(1)/(2) f_(0)T^(2)` `:. v= int dv= int [f_(0) (1-(t)/(T))] dt or, v= f_(0) (t-(t^(2))/(2T))+C` where C is the constant of integration. At t= 0, v= 0 `:. 0 = f_(0) (0-(0)/(2T))+ C rArr C= 0` `:. v= f_(0) (t- (t^(2))/(2T))` If f= 0, then `0 = f_(0) (1- (t)/(T)) rArr t= T` Hence, particle.s velocity in the TIME INTERVAL t=0 and t= T is GIVEN by `v_(x) = underset(t=0)overset(t=T)int dv= underset(t=0)overset(T)int [f_(0) (1-(t)/(T))]dt= (1)/(2) f_(0)T` |
|