A force 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.

Answer: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)