The acceleration experienced by a moving boat after is engine is cut-off, is given by a=-kv3, where k is a constant. If v0 is the magnitude of velocity at cut-off, then the magnitude of the velocity at time t after the cut off is

Question

The acceleration experienced by a moving boat after is engine is cut-off, is given by a=-kv3, where k is a constant. If v0 is the magnitude of velocity at cut-off, then the magnitude of the velocity at time t after the cut off is (A) (v0/2 k t v02) (B) (v0/1 + 2 k t v02) (C) (v0/√1 - 2 k v02) (D) (v0/√1 + 2 k t v02)

Tardigrade Admin · Accepted Answer

Given, acceleration a=-kv3 Initial velocity at cut-off, v1=v0 Initial time of cut-off, t=0 And final time after cut-off, t2=t Again, a=(d v/d t)=-kv3 Or (d v/v3)=-kdt Integrating both sides, with in the condition of motion. ∫ limits v0v (d v/v3)=-∫ limits 0t k d t Or [- (1/2 v2)]v0v=-[k t]0t Or (1/2 v2)-(1/2 v02)=kt Or v=(v0/√1 + 2 k t v02)

Q. The acceleration experienced by a moving boat after is engine is cut-off, is given by $a = - k v^{3},$ where $k$ is a constant. If $v_{0}$ is the magnitude of velocity at cut-off, then the magnitude of the velocity at time $t$ after the cut off is

$\frac{v _{0}}{2 k t v _{0}^{2}}$

$\frac{v _{0}}{1 + 2 k t v _{0}^{2}}$

$\frac{v _{0}}{1 - 2 k v _{0}^{2}}$

$\frac{v _{0}}{1 + 2 k t v _{0}^{2}}$

Q. The acceleration experienced by a moving boat after is engine is cut-off, is given by a=−kv3, where k is a constant. If v0​ is the magnitude of velocity at cut-off, then the magnitude of the velocity at time t after the cut off is

2ktv02​v0​​

1+2ktv02​v0​​

1−2kv02​​v0​​

1+2ktv02​​v0​​

Q. The acceleration experienced by a moving boat after is engine is cut-off, is given by $a = - k v^{3},$ where $k$ is a constant. If $v_{0}$ is the magnitude of velocity at cut-off, then the magnitude of the velocity at time $t$ after the cut off is

$\frac{v _{0}}{2 k t v _{0}^{2}}$

$\frac{v _{0}}{1 + 2 k t v _{0}^{2}}$

$\frac{v _{0}}{1 - 2 k v _{0}^{2}}$

$\frac{v _{0}}{1 + 2 k t v _{0}^{2}}$