JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 13)
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 :-
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 :-
40
32
50
60
Explanation
$$\overrightarrow u $$ = 3.0 $$\widehat i$$
$$\overrightarrow a $$ = $$\left( {6\widehat i + 4\widehat j} \right)$$
$$\overrightarrow S = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$$
x = 3t + $${1 \over 2}6{t^2}$$
= 3t + 3t2 .....(1)
y = $${1 \over 2} \times 4 \times {t^2}$$ = 32
$$ \Rightarrow $$ t = 4 s .... (2)
x = 3 × 4 + 3 × 42 = 12 + 48 = 60 m
$$\overrightarrow a $$ = $$\left( {6\widehat i + 4\widehat j} \right)$$
$$\overrightarrow S = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$$
x = 3t + $${1 \over 2}6{t^2}$$
= 3t + 3t2 .....(1)
y = $${1 \over 2} \times 4 \times {t^2}$$ = 32
$$ \Rightarrow $$ t = 4 s .... (2)
x = 3 × 4 + 3 × 42 = 12 + 48 = 60 m
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