QJEEM20Ph08fb: JEE Main 2020 Physics - Class 11 JEEM20Ph08fb Solution: A particle starts from the origin at t = 0 with an initial…

Question

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A particle starts from the origin at t = 0 with an
initial velocity of 3.0

\hat{i}

m/s and moves in the
x-y plane with a constant acceleration

(6 \hat{i} + 4 \hat{j})

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 :-

Accepted Answer

\vec{u}

= 3.0

\hat{i}

\vec{a}

=

(6 \hat{i} + 4 \hat{j})

\vec{S} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}

x = 3t +

\frac{1}{2} 6 t^{2}

= 3t + 3t² .....(1)

y =

\frac{1}{2} \times 4 \times t^{2}

= 32

\Rightarrow

t = 4 s .... (2)

x = 3 × 4 + 3 × 4² = 12 + 48 = 60 m

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