Question

You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, the car is at r⃗ 0=(3.0i^+2.0j^)mWhat is the x component of the car's position vector at 10 s?What is the y component of the car's position vector at 10 s?What is the x component of the car's acceleration vector at 10 s?What is the y component of the car's acceleration vector at 10 s?

Answer 1

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

$\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s$

is the integral of the velocity.

Integrating $\boldsymbol{v}$ we get the displacement vector $\boldsymbol{d}$:

$\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )$

Now if the initial position if the car is

$\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})$

then the displacement of the car at time $t$ is

$\boldsymbol{d(t)}= \boldsymbol{r+d}$

$\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}$

Now at $t=10s$, we have

$\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m$

The y-component of the car's position vector is 670m/s.

The acceleration vector is the derivative of the velocity vector:

$\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})$

and at $t=10s$ it is

$\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2$

The x-component of the acceleration vector is -3, and the y-component is 40.