Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A spaceship in space sweeps stationary interplanteray dust. As a result, its mass increases at a rate &

A spaceship in space sweeps stationary interplanteray dust. As a result, its mass increases at a rate \frac{dM(t)}{dt}=bv^{2}(t) where v(t) is its instantaeous velocity. The instantaeous accelartion of the satellite is:
Option: 1 \frac{-2bv^{3}}{M(t)}
Option: 2 \frac{-bv^{3}}{M(t)}
Option: 3 -bv^{3}(t)
Option: 4 \frac{-bv^{3}}{2M(t)}

Answers (1)

best_answer

Correct answer : \frac{-bv^{3}}{M(t)}

 

As given

$$\frac{d M(t)}{d t}=b v^{2} (t)$$

As we know,

$$\begin{aligned} F &=\frac{d(M v)}{d t} \\ So, \\ F &= M \frac{d v}{d t}+v \frac{d M}{d t} \end{aligned}$$

$$F=M\left(\frac{d v}{d t}\right)+v\left(b v^{2}\right)$$

\text { We know that the net force is zero. } F=0

\frac{dv}{dt} = a=\left(\frac{-b v^{3}}{M(t)}\right)

Where, M(t) represents Mass as a function of times.

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question