The acceleration experienced by a moving boat after its 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 its 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)  (v0/2ktv02)  (B)  (v0/1+2ktv02)  (C)  (v0/√1-2kv02)  (D)  (v0/√1+2kt 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=(dv/dt)=-kv3 or (dv/v3)=-kdt Integrating both sides, with in the condition ofmotion ∫v0v(dv/v3)=-∫0tk dt or [ -(1/2v2) ]v0v=-[kt]0t or (1/2v2)-(1/2v02)=kt or v=(v0/√1+2kt v02)

Q. The acceleration experienced by a moving boat after its 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 its 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 its 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 :

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

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

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

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