Question

A particle of mass m is located at the origin. It is at rest and in equilibrium. A time-dependent force →F(t) is applied at time t=0, with components Fx(t)=pt and Fy(t)=n+qt where p, q, and n are constants. Find the position →r(t) and velocity →v(t) as functions of time t.

a) →r(t) = (pt²/2, nt + qt²/2) ; →v(t) = (pt, n + qt)

b) →r(t) = (pt²/2, nt + qt) ; →v(t) = (pt, n + qt²/2)

c) →r(t) = (pt²/2, nt + qt²/2) ; →v(t) = (pt, n + qt)

d) →r(t) = (pt²/2, nt + qt) ; →v(t) = (pt, n + qt²/2)

Answer 1

The velocity and acceleration of the particle can be derived from the position function.

Step-by-step explanation:

The position of a particle is given by the function r(t) = (50 m/s)tî - (4.9 m/s²)t²ĵ. To find the velocity and acceleration as functions of time, we need to take the derivative of the position function:

The velocity function is given by v(t) = 50î - 9.8tĵ.

And the acceleration function is given by a(t) = -9.8ĵ.