A particle starts from the origin at t = 0 with an
initial velocity of 3.0 $\widehat i$ m/s and moves in the
x-y plane with a constant acceleration
$\left( {6\widehat i + 4\widehat j} \right)$ m/s2 . The x-coordinate of the
particle at the instant when its y-coordinate is
32 m is D meters. The value of D is :-
Solution
$\overrightarrow u$ = 3.0 $\widehat i$
<br><br>$\overrightarrow a$ = $\left( {6\widehat i + 4\widehat j} \right)$
<br><br>$\overrightarrow S = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$
<br><br>x = 3t + ${1 \over 2}6{t^2}$
<br><br>= 3t + 3t<sup>2</sup> .....(1)
<br><br>y = ${1 \over 2} \times 4 \times {t^2}$ = 32
<br><br>$\Rightarrow$ t = 4 s .... (2)
<br><br>x = 3 × 4 + 3 × 4<sup>2</sup>
= 12 + 48 = 60 m
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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