A force with magnitude 25 N acts directly upward from
the *xy*-plane on an object with mass 5 kg.
The object starts at the origin with initial
velocity **v**(0)
= **i** − **j**. Find
its position function and its speed at time *t*.

r(t)=

absolute value of v(t)=

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(1)Given F= geth 25N and m=5kgAccelelation a=(" force ")/(" mass ")=25//5=5m//s^(2)=5 hat(k)= vec(a)torbxe, force acting in the positive z-dirction{:[a(t)=(d)/(dt)[v(t)]],[int d[v(t)]=int a(t)dt],[v(t)=int5dt hat(k)],[v(t)=5t hat(k)+(a^( bar(i))+bj+c( bar(k)))]:}At t=0,v(0)=0+ai+bj+ck=i-jSo, a=1,b=-1{:[v(t)=i-j+5t bar(k)],[|v(t)|=sqrt(1+1+25t^(2))=sqrt(25t^(2)+2)]:}{:[v(t)=(dx(t))/(dt)=>int dx(t)=int v(t)dt],[=>(t)=int(i- bar(j)+5+ bar(k))dt],[r(t)=t_(i)^( bar(i))-t^( bar(j))+(5)/(2)t^(2) bar(k)+(d_(i)^( bar(i))+( bar(e))( bar(j))+f( bar(k)))]:}At t=0, obrects staits fram origin{:[r(0)=0_(i)^( bar(i))+0 bar(j)+0 bar(k)=(0-0+0)+(di+e bar(j)+f bar(k))],[" so "quad d=0","quad e=0","f=0","]:}than{:[r(t)=t^(-)i-t_(j)^( bar(j))+(5)/(2)t^(2) bar(k)],[r(t)=(:t,-t,(5)/(2)t^(2):)]:} ...