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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2017 · 8 April · Online · Physics · Motion in One Dimension

Which graph corresponds to an object moving with a constant...

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?

Options

  1. AIf g(x)=x^2+x-2 and (1)/(2) g o f(x)=2 x^2-5 x+2 then f(x) is equal to
  2. BIf g(x)=x^2+x-2 and (1)/(2) g o f(x)=2 x^2-5 x+2 then f(x) is equal to
  3. CIf g(x)=x^2+x-2 and (1)/(2) g o f(x)=2 x^2-5 x+2 then f(x) is equal to
  4. DIf g(x)=x^2+x-2 and (1)/(2) g o f(x)=2 x^2-5 x+2 then f(x) is equal to

Correct answer

C. If g(x)=x^2+x-2 and (1)/(2) g o f(x)=2 x^2-5 x+2 then f(x) is equal to

Step-by-step solution

Let acceleration be aa.
Given that a=−Ca = - C
We know that, a=a =vdvdx=−C\dfrac{v d v}{d x} = - C
vdv=−Cdxv d v = - C d x

Integrating the above equation,
v22=−Cx+k\dfrac{{v}^{2}}{2} = - Cx + k.

Thus, the velocity-distance graph is

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?

The distance cover by the object is x=−v22C+kCx = - \dfrac{{v}^{2}}{2C} + \dfrac{k}{C}.

Thus, with the increase in distance, velocity decreases.

Now, a=−C=dvdta = - C = \dfrac{d v}{d t}

⇒dv=−Cdt⇒v=−Ct+u,(straight line)⇒ d v = - C d t ⇒ v = - Ct + u, \left(\text{straight line}\right)

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