A particle starts from the origin at t = 0 with an initial velocity of 3.0 hati m / s and moves in the x-y plane wit h a constant acceleration (6.0 hati+4.0 hatj)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 :

Question

A particle starts from the origin at t = 0 with an initial velocity of 3.0 hati m / s and moves in the x-y plane wit h a constant acceleration (6.0 hati+4.0 hatj)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 : (A) 60 (B) 50 (C) 32 (D) 40

Tardigrade Admin · Accepted Answer

x = uxt +(1/2)axt2 y = uyt +(1/2)ayt2 32 = 0× t +(1/2)(4)(t)2 t2 = 16 t = 4 sec x = 3×4+(1/2)×6×42 = 12 + 48 = 60 m

$60$

$50$

$32$

$40$

$x = u_{x} t + \frac{1}{2} a_{x} t^{2}$
$y = u_{y} t + \frac{1}{2} a_{y} t^{2}$
$32 = 0 \times t + \frac{1}{2} (4) (t)^{2}$
$t^{2} = 16$
$t = 4 sec < b r / > x = 3 \times 4 + \frac{1}{2} \times 6 \times 4^{2}$
$= 12 + 48 = 60 m$

Q. A particle starts from the origin at t=0 with an initial velocity of 3.0i^m/s and moves in the x−y plane wit h a constant acceleration (6.0i^+4.0j^​)m/s2. The x -coordinate of the particle at the instant when its y- coordinate is 32m is D meters. The value of D is :

60

50

32

40

x=ux​t+21​ax​t2 y=uy​t+21​ay​t2 32=0×t+21​(4)(t)2 t2=16 t=4sec<br/>x=3×4+21​×6×42 =12+48=60m

$60$

$50$

$32$

$40$

$x = u_{x} t + \frac{1}{2} a_{x} t^{2}$
$y = u_{y} t + \frac{1}{2} a_{y} t^{2}$
$32 = 0 \times t + \frac{1}{2} (4) (t)^{2}$
$t^{2} = 16$
$t = 4 sec < b r / > x = 3 \times 4 + \frac{1}{2} \times 6 \times 4^{2}$
$= 12 + 48 = 60 m$