After 5 seconds velocity is given by
v → = u → + 5 a → \overset{\to}{v}=\overset{\to}{u}\ +\ 5\overset{\to}{a} v → = u → + 5 a →
Talking about components
v x = u x + 5 a x v y = u y + 5 a y v_x=u_x\ +\ 5a_x\\v_y=u_y\ +\ 5a_y v x = u x + 5 a x v y = u y + 5 a y
v x = 4.69 cos ( 40 ° ) + 5 × 1.85 cos ( 200 ° ) v y = 4.69 sin ( 40 ° ) + 5 × 1.85 sin ( 200 ° ) v_x=4.69\cos(40\degree)+5\times1.85\cos(200\degree)\\v_y=4.69\sin(40\degree)+5\times 1.85\sin(200\degree) v x = 4.69 cos ( 40° ) + 5 × 1.85 cos ( 200° ) v y = 4.69 sin ( 40° ) + 5 × 1.85 sin ( 200° )
v x = − 5.098 v y = − 0.149 ∣ v ∣ = v x 2 + v y 2 = 5.1 m s e c − 1 v_x=-5.098\\v_y=-0.149\\|v|=\sqrt{v_x^2+v_y^2}=5.1\ m\ sec^{-1} v x = − 5.098 v y = − 0.149 ∣ v ∣ = v x 2 + v y 2 = 5.1 m se c − 1
tan ( θ ) = 0.149 5.098 = 0.0292 θ = 1.67 s i n c e b o t h c o m p o n e n t a r e n e g a t i v e s o a n g l e w i l l b e 180 + 1.67 = 181.67 ° \tan(\theta)=\frac{0.149}{5.098}=0.0292\\\theta=1.67\\since \ both\ component\ are\ negative\ so angle \ will\ be\ 180+1.67=181.67\degree tan ( θ ) = 5.098 0.149 = 0.0292 θ = 1.67 s in ce b o t h co m p o n e n t a re n e g a t i v e so an g l e w i ll b e 180 + 1.67 = 181.67° So velocity after 5 seconds is elapsed will be
5.1 m s e c − 1 5.1\ m\ sec^{-1} 5.1 m se c − 1 at an angle of 181.67 ° 181.67\degree 181.67°
Or in term of vector component
v → = − 5.098 i ^ − 0.149 j ^ \overset{\to}{v}=-5.098\hat{i}-0.149\hat{j} v → = − 5.098 i ^ − 0.149 j ^
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