Motion in a Plane Easy Questions and Answers

F^{^→} = 6t î + 4t ĵ
F_x = 6t , F_y = 4t

a_x =	6t	= 2t , a_y =	4t
	3		3

v_x = 0 + 2t . t = 18 , for t = 3s

v_y = 0 +	4	t.t = 12 for t = 3s
	3

Velocity → 18î + 12ĵ

F^{^→} = 6t î + 4t ĵ
F_x = 6t , F_y = 4t

a_x =	6t	= 2t , a_y =	4t
	3		3

v_x = 0 + 2t . t = 18 , for t = 3s

v_y = 0 +	4	t.t = 12 for t = 3s
	3

Velocity → 18î + 12ĵ

Let after 1 sec angle become 60°. When the end A moves by 10 m left, the end B moves upward by BB′ = 10 × √3 = 10 × 1.73 = 17.3 m / s

Let after 1 sec angle become 60°. When the end A moves by 10 m left, the end B moves upward by BB′ = 10 × √3 = 10 × 1.73 = 17.3 m / s

r^{^→} = (a cos ωt)î + (a sin ωt)ĵ

v^{^→} =	dr^{^→}	=	d	{ (a cos ωt)î + (a sin ωt)ĵ }
	dt		dt

= (-aω sin ωt)î + (aω cos ωt)ĵ
= ω [ (-a sin ωt)î + (a cos ωt)ĵ ]

Slope of position vector =	a sin ωt	= tan ωt
	a cos ωt

& slope of velocity vector =	-a cos ωt	=	-1
	a sin ωt		tan ωt

∴ velocity is perpendicular to the displacement.

r^{^→} = (a cos ωt)î + (a sin ωt)ĵ

v^{^→} =	dr^{^→}	=	d	{ (a cos ωt)î + (a sin ωt)ĵ }
	dt		dt

= (-aω sin ωt)î + (aω cos ωt)ĵ
= ω [ (-a sin ωt)î + (a cos ωt)ĵ ]

Slope of position vector =	a sin ωt	= tan ωt
	a cos ωt

& slope of velocity vector =	-a cos ωt	=	-1
	a sin ωt		tan ωt

∴ velocity is perpendicular to the displacement.

Speed of the bullet (v) = 1000 m/s and horizontal distance of the target (s) = 100 m.

Time taken to cover the horizontal distance (t) =	100	= 0.1 sec
	1000

During this time, the bullet will fall down vertically due to gravitational acceleration.

∴ height = ut +	1	gt²
	2

= (0 × 0.1) +	1	10(0.1)² = 0.05 m = 5 cm
	2

Speed of the bullet (v) = 1000 m/s and horizontal distance of the target (s) = 100 m.

Time taken to cover the horizontal distance (t) =	100	= 0.1 sec
	1000

During this time, the bullet will fall down vertically due to gravitational acceleration.

∴ height = ut +	1	gt²
	2

= (0 × 0.1) +	1	10(0.1)² = 0.05 m = 5 cm
	2

Motion in a Plane