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JIPMER 2008
Exam
JIPMER
Year
2008
Subject
Physics
Chapter
Motion in One Dimension

JIPMER 2008 · Physics · Motion in One Dimension

The displacement of particle is given by x=a_0+(a_1 t)/(2)-(a_2...

The displacement of particle is given by x=a0+a1t2−a2t23x=a_0+\frac{a_1 t}{2}-\frac{a_2 t^2}{3} What is its acceleration?

Options

  1. A2a23\frac{2 a_2}{3}
  2. B−2a23-\frac{2 a_2}{3}
  3. Ca2a_2
  4. DZero

Correct answer

B. −2a23-\frac{2 a_2}{3}

Step-by-step solution

Key Idea : Acceleration is the rate of change of velocity and velocity is the rate of change of displacement. The displacement equation is given by x=a0+a1t2−a2t23x=a_0+\frac{a_1 t}{2}-\frac{a_2 t^2}{3} Velocity = rate of change of displacement ie, v=dxdt\quad v=\frac{d x}{d t}=ddt(a0+a1t2−a2t23)=0+a12−2a2t3=a12−2a2t3\begin{aligned} & =\frac{d}{d t}\left(a_0+\frac{a_1 t}{2}-\frac{a_2 t^2}{3}\right) \\ & =0+\frac{a_1}{2}-\frac{2 a_2 t}{3} \\ & =\frac{a_1}{2}-\frac{2 a_2 t}{3}\end{aligned}Acceleration = rate of change of velocity $i e$,\begin{aligned} a & =\frac{d v}{d t} \\ & =\frac{d}{d t}\left(\frac{a_1}{2}-\frac{2 a_2}{3} t\right) \\ & =0-\frac{2 a_2}{3} \\ & =-\frac{2 a_2}{3}\end{aligned}$$
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