Question

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?

Answer 1

given

y = 33 m

velocity of magnitude v = 3.5 m/s

acceleration of magnitude a = 0.46 m/s²

the horizontal component is

x_A = x_B

V_A t = 1/2 a sin_t²

V_A = 1/2 a sin_t  

sin = 2 V_A / a t

y_B = 33 m

1/2 a cos_t² = 33

cos = 66 / a t²

sin = ( 1 - cos² )^1/2

= ( 1 - (66 / a t²)² )^1/2

V_A = 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 (2v_A / at )

= sin^-1 (2 x 3.5 / 0.46 x 17.324 )

= 61.44^o

so the angle θ between a→ and the positive direction of the y axis would result in a collision is = 61.44^o