Question

An object experiences a constant acceleration of 2.00 m/s² along the -x axis for 2.70 s, attaining a velocity of 18.0 m/s in a direction 47∘∘ from the +x axis.

1)

Calculate the magnitude of the initial velocity vector of the object. (Express your answer to two significant figures.)

2)

Calculate the direction of the initial velocity vector of the object. Find the angle this vector makes with respect to the +x axis. Use value from -180 to +180. (Express your answer to two significant figures.)

Answer 1

The final velocity vector can be written as

v = 18 cos 47^o i + 18 sin 47^o j

v = 12.27 i + 13.16 j

Let the initial velcoity vectro be

u = u_x i + u_y j

a_x = - 2

t = 2.7 s

v_x = 12.27 (x- component of final velocity)

so

v_x = u_x + a_xt

12.27 = u_x + (-2)(2.7)

12.27 = u_x - 5.4

u_x = 17.67

Since there is no acceleration in y - direction the y - component of velocity will not change

so u_y = v_y

u_y = 13.16

so initial velocity vector will be

u = 17.67 i + 13.16 j

2) The direction of the vector will be

The angle made by initial vector with + x-axis is 37^o