Question

A body of mass 2 kg at O has an initial velocity of 3m/s along OE and it is subjected to a force of 4N perpendicular to OE the distance covered in 4 seconds?

Answer 1

Perpendicular acceleration:
F = ma
a = 4 / 2 = 2 m/s²

Perpendicular distance:

s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
s = 16 m

Horizontal distance:
s = ut
= 3 x 4
= 12 m

Total distance = √(12² + 16²)
= 20 m.