Question

1. An object of 3.00 kg moves in the Cartesian plane with its coordinates (x, y) known as ? =5?^2 - 1 and ? = 3?^3 + 2  where x and y are in meters and t is in seconds. Determine the force acting in the object at t = 2.00 s.

2.Three forces acting on an object are provided by ?⃗1 = (-2.00? + 2.00?) ?, ?⃗2 = (5.00?-3.00?) ? and ?⃗3 = (-45.0?) ?. If the acceleration is 3.75 m /s2 ,determine

(a) the mass of the object and

(b) the distance that runs at 10 seconds, if it was at rest initially.

Answer 1

1. Given, and

and

So, acceleration in x direction = 10

acceleration in y direction = 18t

SO, net acceleration = 10i + 18tj Ans

2. Total force in x direction = -42i

total force in y direction = -1j

So net force = (42^2 + 1)^.5 = 42.01

So, net acceleration = 42.01/m = 3.75

m = 42.01/3.75 = 11.2 kg

distance s = 1/2at2 = 1/2x3.75x100 = 187.5 m