Easy MCQ +4 / -1 PYQ · JEE Mains 2020

Starting from the origin at time t = 0, with initial velocity 5$\widehat j$ ms-1 , a particle moves in the x-y plane with a constant acceleration of $\left( {10\widehat i + 4\widehat j} \right)$ ms-2. At time t, its coordinates are (20 m, y0 m). The values of t and y0 are, respectively:

  1. A 5s and 25 m
  2. B 2s and 18 m Correct answer
  3. C 2s and 24 m
  4. D 4s and 52 m

Solution

$y = {u_y}t + {1 \over 2}{a_y}{t^2}$<br><br>$y = 5t + {1 \over 2}(4){t^2}$<br><br>$y = 5t + 2{t^2}$<br><br>and $x = 0(t) + {1 \over 2}(10)({t^2}) = 20$<br><br>$t = 2s$<br><br>$\Rightarrow y = 10 + 8 = 18m$

About this question

Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line

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