Answer :
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
[tex]\vec{r}= 2+2t^2, 6+t^3,0[/tex]
at t=2s, the position vector is
[tex]\vec{r}= 10, 14,0[/tex]
Therefore, the the distance between the car and the girl is
[tex]s= \sqrt{(10-0)^2+(14-0)^2+(0-0)^2)}\[/tex]
s = 17.20
(b)
The position of the car at t = 0s is [tex]\vec{r}_0 = 2,6,0[/tex]
The position of the car at t = 2s is [tex]\vec{r}_2 = 10,14,0[/tex]
The distance of the car traveled in the interval from t=0s to t=2 s is as follows:
[tex]s_{02}= \sqrt{(10-2)^2+(14-6)^2+(0-0)^2)} \\ \\ s_{02} = 11.31[/tex]
(c)
The position vector of the car at time 't' secs is
[tex]\vec{r}= 2+2t^2, 6+t^3,0[/tex]
The velocity of the car is
[tex]\vec{v}=\dfrac{d\vec{r}}{dt}= 4t, 3t^2,0[/tex]
the direction of the car's velocity at t = 2s is going to be
[tex]\vec{v}\mid _t=2 8, 12,0[/tex]
Thus; The speed of the car is
[tex]v_{t=2}= \sqrt{8^2+12^2+0^2} \\ \\ v_{t=2}= 14.42[/tex]
(d) the car's acceleration is:
[tex]\vec{a}=\frac{d\vec{v}}{dt}= 4, 6t,0[/tex]
The magnitude of car's acceleration at t=2s is
[tex]\mid \vec{a}\mid _{t=2}=\sqrt{4^2+12^2+0^2} \\ \\ \mid \vec{a}\mid _{t=2}= 12.65[/tex]