Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

An object moving with a speed of 6.25 \: m\: s^{-1}, is decelerated at a rate given by \frac{d\upsilon }{dt}= -2.5\sqrt{\upsilon }

where \upsilon is the instantaneous speed. The time taken by the object, to come to rest, would be

  • Option 1)

    1\: s

  • Option 2)

    2\: s

  • Option 3)

    4\: s

  • Option 4)

    8\: s

 

Answers (1)

best_answer

As we learnt in

Uniform velocity,Non uniform Velocity -

Uniform velocity

If equal displacement are covered in equal intervals of time.

Non uniform velocity

If equal displacement are covered in unequal intervals of time.

- wherein

More to know :

For uniform motion along a straight line, the average and instantaneous velocities have the same values.

 

   \frac{dv}{dt}=-2.5\sqrt{v}\ \;or\ \;\frac{1}{\sqrt{v}}dv = -2.5 dt

\int_{v1}^{v2} \frac{dv}{\sqrt{v}} = -2.5 \int_{0}^{t}dt

v_{1} = 6.25\ ms^{-1}

v_{2} = 0

2 [v^{1/2}]^0_{6.25}= -2.5 t \Rightarrow t=\frac{-2\times (6.25)^{1/2}}{-2.5}

t=2sec


Option 1)

1\: s

Incorrect

Option 2)

2\: s

Correct

Option 3)

4\: s

Incorrect

Option 4)

8\: s

Incorrect

Posted by

Aadil

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

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today
anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

Latest Question