Aforce with magnitude 20 n acts directly upward from the xy-plane on an object with mass 4 kg. the object starts at the origin with initial velocity vs0d − i 2 j. find its position function and its speed at time t.

position function at time,t=r(t)=ti-tj+5t^2K+C

Speed at time,t=square root(2+25t^2)

Step-by-step explanation: Force= mass× acceleration =F=ma

a=f/m

Given: force=20N ,Mass=4kg

a=20/4=5m/s

a(t)=5K. Using integration

V(t)=5tK+C

Where C is constant of integration.

Given:v(0)=i-j

V(0)=5(0)+K

Zi-j=0+C integrating to get the position function gives:

r(t)=ti-tj+5t^2 K +C

C=0 because object started at origin

Speed is magnitude of velocity

V(t)=i-j+5t K

/v(t)/÷ squareroot(1^2+(-1)^2)+5t^2

Speed = squareroot (2+25t^2)