Răspuns:
Ex1 r(t)=(2t²-1)i+5t²j
Ecuatia traiectoriei
{x(t)=2t²-1
{y(t)=5t²
viteza
v=dr/dt
{vx=(2t²-1) `=4t
{vy=(5t²)`=10t
acceleratia
{ax =vx `(t)=(4t) `=4
{ay=vy `(t)=(10t) `=10
F=ma
a=√(ax²+ay²)=√(4²+10²)=√116=2√29
F=1*2√29=2√29
Mai iti fac unul
Ex4
r(t)=(8t⁴+3)i+2t²k m=2kg
Ecuatiile traiectoriei
{x(t)= 8t⁴+3
{y(t)=0
{z(t0+2t²
viteza v=dr/dt
{vx(t)=(8t⁴+3)`=32t³
{vy(t)=0 `=0
{z(t)=2t²) `=4t
Aceleratia
a=dv/dt
{ax=vx `(t)=(32t³) `=96t²
{ay=0 `=0
{az= vz `(t)=4
F=ma
ax(1)=96*1²=96
ay(1)=0
ax=4
a=√(96²+0²+4²)= ....
F=.....
Explicație: