A 2.0 kg frictionless puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.5 s?

the distance moved by the puck after 2.5 s is 7.8 m

Explanation:

Given;

mass of the puck, m = 2 kg

initial velocity of the puck, u = 0

applied force, F = 5 N

time of motion, t = 1.5 s

Acceleration of the puck is calculated from Newton's second law of motion;

F = ma

a = F/m

a = 5/2

a = 2.5 m/s²

The distance moved by the puck after 2.5 s is calculated as;

s = ut + ¹/₂at

s = 0 + ¹/₂at²

s = ¹/₂at²

s = 0.5 x 2.5 x (2.5)²

s = 7.8 m

Therefore, the distance moved by the puck after 2.5 s is 7.8 m

prolly a beacuse the momentuem the car has increases while going down hill

. west