Q108 / 200 · NEET 2008 (Screening) · Physics

A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 \mathrm{~ms}^{-1}$ to $20 \mathrm{~ms}^{-1}$ while passing through a distance $135 \mathrm{~m}$ in $t$ second. The value of $t$ is

  • A
    10
  • B
    1.8
  • C
    12
  • Correct answer
    D
    9

Solution

Key Idea : The problem requires kinematics equations of motion. Let $u$ and $v$ be the first and final velocities of particle and $a \& s$ be the constant acceleration and distance covered by it. From third equation of motion $\begin{array}{l} y^2 =u^2+2 a s \\ \Rightarrow (20)^2 =(10)^2+2 a \times 135 \\ \text {or } a =\frac{300}{2 \times 135}=\frac{10}{9} \mathrm{~ms}^{-2} \end{array}$ Now using first equation of motion, $\begin{aligned} & v=u+a t \\ & \text {or } t=\frac{v-u}{a}=\frac{20-10}{(10 / 9)}=\frac{10 \times 9}{10}=9 \mathrm{~s} \end{aligned}$

Hence, the correct answer is option: 'D'.

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