In the figure, particle A moves along the line y = 33 m with a constant velocity v→ of magnitude 3.5 m/s and directed parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a→ of magnitude 0.46 m/s2. What angle θ between a→ and the positive direction of the y axis would result in a collision?
given
y = 33 m
velocity of magnitude v = 3.5 m/s
acceleration of magnitude a = 0.46 m/s2
the horizontal component is
xA = xB
VA t = 1/2 a sint2
VA = 1/2 a sint
sin = 2 VA / a t
yB = 33 m
1/2 a cost2 = 33
cos = 66 / a t2
sin = ( 1 - cos2 )1/2
= ( 1 - (66 / a t2)2 )1/2
VA = 1/2 at ( 1 -(66 / a t2)2 )1/2
3.5 = 1/2 x 0.46 t ( 1 -(66 / 0.46 t2)2 )1/2
15.21 = t ( 1 -(143.47 t2)2 )1/2
15.21 = t ( 1 - 20583.64/t4 )1/2
231.557 = t2 ( 1 - 20583.64/t4 )
t2 - 20583.64/t2 - 231.557 = 0
t4 - 231.557 t2 - 20583.64 = 0
t2 = 300.132
t = 17.324 sec
= sin-1 (2vA / at )
= sin-1 (2 x 3.5 / 0.46 x 17.324 )
= 61.44o
so the angle θ between a→ and the positive direction of the y axis would result in a collision is = 61.44o
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