Question

The motion of a particle is defined by the equations x = (2t + t?) m and y = (t2) m, where t is in seconds. Determine the normal and tangential components of the particle's velocity and acceleration when t = 2 s. Select one 0 a. Vt= 6.0 m/s. Vn= 4.0 m/s, at= 2.0 m/s2, an= 2.0 m/s2 b. Vt= 1.55 m/s. Vn= 6.2 m/s at= 5.3 m/s2 an= 3.2 m/s2 c. Vt= 5.3 m/s. Vn= 3.2 m/s at= 6.2 m/s2 an= 1.55 m/s2 d. Vt= 2.79 m/s. Vn= 4.6 m/s at= 7.33 m/s2 an= 0.44 m/s2 e. Vt= 7.21 m/s. Vn= 0 m/s at= 2.77 m/s2 an= 0.55 m/s2

Answer 1

To find the tangential and normal components, we need to find the slope of the tangent to the trajectory curve at that instant. But for finding the slope to the curve, we need the equation of the curve. y 0 х trajectory For this take the equation for y: y = ť .:t= Vy (since t > 0) Substitute this in equation for x: x = 2t + t2 = 27y + y Now use this to find the angle of the tangent with the y-axis 8: tan = dx dy d (27y + y) dy 1 1 = 2 x X+1 2 Ty 1 + 1 Vy 1 ::= tan-1 +1 Substitute t = 2 s in the equation for y: y = 22 = 4 m Substitute this to find 0:

1 0 = tan-1 n-G+1) :: 56.31° Now, let's find the components of velocity in x and y directions: dx Vx = dt d (2t + t2) dt = 2 + 2t Substitute t= 2 s: Vx = 2 + 2 x 2 = 6 m/s And, Vy dy dt d dt = 2t = (2) Substitute t= 2 s: Vy = 4 m/s Next, find components of acceleration in x and y directions: dvx az = dt d (2 + 2t) dt = 2 m/s2 = And, ау dvy dt d (2t) dt 2 m/s2 Now, find the normal and tangential components of the velocity: Vi = Vx sin @ + vycos 0 = 6 sin 56.31° + 4 cos 56.31° :: V4 ~ 7.21 m/s2 And, Vn = -V, cos 0 + v, sin 0 = -6 cos 56.31° + 4 sin 56.31°

| Pa = 0 m/s2 Next, find the normal and tangential components of the acceleration: a cos ax sin 0 + ау = 2 sin 56.31° + 2 cos 56.31° :: Qį = 2.77 m/s2 And, Añ = -ax cos O + ау sin e = -2 cos 56.31° + 2 sin 56.31° :: An = 0.55 m/s2 Therefore, option e. is the correct option