Question
Question: The acceleration experienced by a moving boat after its engine cut-off, is given by: \[a = - k{v^3}\...
The acceleration experienced by a moving boat after its engine 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 cut-off is:-
A. 2ktv02v0
B. 1+ktv02v0
C. 1−2ktv02v0
D. 1+2ktv02v0
Solution
Initial velocity is the velocity at which time is zero. Acceleration is the time derivative of the velocity of the object. Integrating variable separable differential equations gives the value of unknown.
Formula used:
We have a relationship that is, the relation between acceleration and velocity, that is,
a=dtdv …………..(1)
Formula used:
Here we are asked to find the final velocity where initial velocity is given as v0 and acceleration is given as a=−kv3. Substituting this value into (1) equation we get,
dtdv=−kv3 arranging in variable separable form,
v3dv=−kdt integrating on both sides gives,
∫v3dv=−∫kdt
2v2−1=−kt+C………….(2), to find the value of integrating constant C , we want to find the velocity at the time t=0 , that is at initial velocity.
2v02−1=C…………(3)
Hence substituting (3) in (2) gives,
2v2−1=−kt−2v021
rearranging and taking common −1 from both sides we will get,
2v2=kt+2v0211
again rearranging we will get as,
2v2=1+2ktv022v02
canceling common 2 and taking square roots we get the final velocity as,
v=1+2ktv02v0
Hence, option D is correct.
Note:
Here acceleration is negative. Negative acceleration is known as deceleration or retardation
Retardation acts in the opposite direction of motion and reduces the velocity.
Partial derivatives should be used when acceleration depends on more than one variable. But here acceleration is only the function of velocity.