Answer:
The y-component of the car's position vector is 670m/s.
The x-component of the acceleration vector is -3, and the y-component is 40.
Explanation:
The displacement vector of the car with velocity
[tex]\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s[/tex]
is the integral of the velocity.
Integrating [tex]\boldsymbol{v}[/tex] we get the displacement vector [tex]\boldsymbol{d}[/tex]:
[tex]\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )[/tex]
Now if the initial position if the car is
[tex]\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})[/tex]
then the displacement of the car at time [tex]t[/tex] is
[tex]\boldsymbol{d(t)}= \boldsymbol{r+d}[/tex]
[tex]\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}[/tex]
Now at [tex]t=10s[/tex], we have
[tex]\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m[/tex]
The y-component of the car's position vector is 670m/s.
The acceleration vector is the derivative of the velocity vector:
[tex]\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})[/tex]
and at [tex]t=10s[/tex] it is
[tex]\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2[/tex]
The x-component of the acceleration vector is -3, and the y-component is 40.
