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Motion in A Straight Line question

2019 · 9 Apr · Shift 2 · Q57
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  5. /2019 · 9 Apr · Shift 2 · Q57

Motion in A Straight Line question

2019 · 9 Apr · Shift 2 · Q57

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
  1. A
    a+b2ca + {{{b^2}} \over {c}}a+cb2​
  2. B
    a+b24ca + {{{b^2}} \over {4c}}a+4cb2​
  3. C
    a+b23ca + {{{b^2}} \over {3c}}a+3cb2​
  4. D
    a+b22ca + {{{b^2}} \over {2c}}a+2cb2​
View written solutionFree

Correct answer: C

  1. Given position function

    x(t)=at+bt2−ct3x(t)=at+bt^2-ct^3x(t)=at+bt2−ct3

  2. Find velocity

    Velocity is the first derivative of position:

    v(t)=dxdt=a+2bt−3ct2v(t)=\frac{dx}{dt}=a+2bt-3ct^2v(t)=dtdx​=a+2bt−3ct2

  3. Find acceleration

    Acceleration is the derivative of velocity:

    ax(t)=dvdt=2b−6cta_x(t)=\frac{dv}{dt}=2b-6ctax​(t)=dtdv​=2b−6ct

  4. Condition for zero acceleration

    Set acceleration equal to zero:

    2b−6ct=02b-6ct=02b−6ct=0

    6ct=2b6ct=2b6ct=2b

    t=b3ct=\frac{b}{3c}t=3cb​

  5. Find velocity at this instant

    Substitute t=b3ct=\frac{b}{3c}t=3cb​ into

    v(t)=a+2bt−3ct2v(t)=a+2bt-3ct^2v(t)=a+2bt−3ct2

    So,

    v=a+2b(b3c)−3c(b3c)2v=a+2b\left(\frac{b}{3c}\right)-3c\left(\frac{b}{3c}\right)^2v=a+2b(3cb​)−3c(3cb​)2

    v=a+2b23c−3c⋅b29c2v=a+\frac{2b^2}{3c}-3c\cdot \frac{b^2}{9c^2}v=a+3c2b2​−3c⋅9c2b2​

    v=a+2b23c−b23cv=a+\frac{2b^2}{3c}-\frac{b^2}{3c}v=a+3c2b2​−3cb2​

    v=a+b23cv=a+\frac{b^2}{3c}v=a+3cb2​

  6. Match with options

    This corresponds to:

    Option C: a+b23ca+\frac{b^2}{3c}a+3cb2​

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