Question

A girl operates a radio-controlled model car in a vacant parking lot. The girl's position is at the origin of the xy coordinate axes, and the surface of the parking lot lies in the x-y plane. The motion of the car is defined by the position vector r = (2 + 2t2)i + (6 + t3 )j where r and t are expressed in meters and seconds, respectively. Determine (a) the distance between the car and the girl when t = 2s, (b) the distance the car traveled in the interval from t = 0 to t = 2s, (c) the speed and direction of the car's velocity at t = 2s, (d) the magnitude of the car's acceleration at t = 2s.

Answer 1

Answer:

a) 17.20

b) 11.31

c) 14.42

d) 12.65

Explanation:

(a)

The girl is at the origin of the x,y coordinates  (i.e 0,0,0  )

the position vector of the car at time 't' secs is

$\vec{r}= 2+2t^2, 6+t^3,0$

at t=2s,  the position vector is

$\vec{r}= 10, 14,0$

Therefore, the the distance between the car and the girl is

$s= \sqrt{(10-0)^2+(14-0)^2+(0-0)^2)}$

s = 17.20

(b)

The position of the car at  t = 0s is $\vec{r}_0 = 2,6,0$

The position of the car at t = 2s is $\vec{r}_2 = 10,14,0$

The distance of the car traveled in the interval from t=0s to t=2 s is as follows:

$s_{02}= \sqrt{(10-2)^2+(14-6)^2+(0-0)^2)} \\ \\ s_{02} = 11.31$

(c)

The position vector of the car at time 't' secs is

$\vec{r}= 2+2t^2, 6+t^3,0$

The velocity of the car is

$\vec{v}=\dfrac{d\vec{r}}{dt}= 4t, 3t^2,0$

the direction of the car's velocity at t = 2s is  going to be

$\vec{v}\mid _t=2 8, 12,0$

Thus; The speed of the car is

$v_{t=2}= \sqrt{8^2+12^2+0^2} \\ \\ v_{t=2}= 14.42$

(d)    the car's acceleration is:

$\vec{a}=\frac{d\vec{v}}{dt}= 4, 6t,0$

The magnitude of car's acceleration at t=2s is

$\mid \vec{a}\mid _{t=2}=\sqrt{4^2+12^2+0^2} \\ \\ \mid \vec{a}\mid _{t=2}= 12.65$