PHYSICS HOMEWORK

Question

Each of the following vectors is given in terms of its x- and y-components. Draw the vector, label an angle that specifies the vectors direction, then find the vectors magnitude and direction.

a) V_x= 20 m/s, V_y = 40 m/s

b) a_x =2.0 m/s^2, a_y= -6.0 m/s²

Answer 1

Hi Jonathan G.,

In two dimensions, the magnitude of the velocity (speed) is equal to the square root of the sum of its squared components, |v| = [ (v_x)² + (v_y)² ]^1/2 and the direction is found by the inverse tangent of y component divided by the x component, ∠v = tan^-1 (v_y/v_x).

a) |v| = [ (20)² + (40)² ]^1/2 = 20√5 = 44.7 or 45 m/s rounded.

∠v = tan^-1(40/20) = 63.4^o or 63^o rounded.

The velocity vector v has a magnitude of 45 m/s in the direction above the +x axis 63^o towards the +y axis.

b) See if you can do the acceleration a. |a| = [ (a_x)² + (a_y)² ]^1/2 and ∠a = tan^-1 (a_y/a_x)

The answers are |a| = 6.3 m/s² and ∠a = -72^o (clockwise below the +x axis).

Have fun! I hope this helps, Joe.