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)

she played for 18.75 hours.

step-by-step explanation:

if she played 2/4 of a hour on tuesday. and there is 60 minutes in an hour and 60 divided by 4 is 7.5. and she played 2/4 of an hour on thursday then again 60 divided by 4 is 7.5, you add 7.5+7.5=56.25. now you have to do 1/4 of an hour, it equals 3.75. now you have to add them all up. 3.75+7.5+7.5=18.75

i hope this and i hope you get it right. it took me a while to get this done for you. : )