Answer to Question #111630 in Vector Calculus for Felix Coleman

Question

1. The position vector r of a missile at any time t is given by
r = ti + (8t − t
2
)j,
where i and j are unit vectors in the horizontal and upward vertical directions. Find its
(a) average velocity from t = 2 to t = 3,
(b) velocity, speed and direction of motion when t = 4,
(c) position vector when it is moving parallel to the vector i − 2j,
(d) acceleration vector.
2. The position vector r of a moving particle at time t after the start of the motion is given by
r = 5(1 + 4t)i + 5(19 + 2t − t
2
)j.
Find the initital velocity of the particle. At time t = T, the particle is moving at right angles to its initial
direction of motion. Find the value of T and the distance of the particle from its initial position at this
time.

Answer 1

(a) average velocity from t = 2 to t = 3,

$r = ti + (8t − t^2)j$

$average\phantom{i}velocity=\frac{r(3)-r(2)}{3-2}\\$

$r(3) = 3i + (8(3) − (3)^2)j\\r(3) = 3i + 15j\\r(2)= 2i + (8(2) −(2)^2)j\\r(2) = 2i + 12j$

hence

$average\phantom{i}velocity=\frac{r(3)-r(2)}{3-2}\\average\phantom{i}velocity=\frac{(3i+15j)-(2i+12j)}{3-2}\\average\phantom{i}velocity=\frac{i+3j}{1}\\average\phantom{i}velocity=i+3j$

(b) velocity, speed and direction of motion when t = 4,

velocity function

$r = ti + (8t − t^2)j$

$velocity=\frac{dr}{dt}\\velocity=\frac{dr}{dt}=i+(8-2t)j$

hence at t=4

$velocity=i+0j\implies velocity=i$

for speed

$speed=|\frac{dr}{dt}|\\\frac{dr}{dt}=i$

hence speed

$speed=\sqrt{1^2}\implies speed=1$

direction of motion

$direction=\frac{v(t)}{|v(t)|}\\direction=\frac{i}{1}\\direction=i$

$tan\theta=\frac{y-component}{x-component}\phantom{i}of\phantom{i}the\phantom{i}velocity\\ \theta=tan^{-1}(\frac{0}{1})\\\theta=0^{0}$

both methods implies the particle is moving along the x-axis

c.) position vector when it is moving parallel to the vector i − 2j

$\begin{pmatrix} t \\ 8t-t^2 \end{pmatrix}=k\begin{pmatrix} 1 \\ -2 \end{pmatrix}\\\implies t=k\\8t-t^2=-2k\\$

solving the two equations simultaneously by substitution

$8t-t^2=-2t\\hence\\t=10 \\\therefore r=10i-20j$

d.) acceleration vector

$acceleration=\frac{dv}{dt}\\\frac{dv}{dt}=-2j$

2.) initial velocity

$r = 5(1 + 4t)i + 5(19 + 2t − t^2)j. \\velocity=\frac{dr}{dt}\\\frac{dr}{dt}=20i+5(2-2t)j\\velocity=20i+5(2-2t)j \\at\phantom{i}t=0$

initial velocity

$v=20i-10tj$

value of T

$r = 5(1 + 4t)i + 5(19 + 2t − t^2)j$

two vectors are said to be perpendicular if their dot product is equal to zero

initial displacement is when t=0

$t=0\\r = 5i + 95j$

at t=T

$r = 5(1 + 4T)i + 5(19 + 2T − T^2)j$

then

$(5(1 + 4T)i + 5(19 + 2t − T^2)j).(5i+95j)=0\\(5(1 + 4T)5)+(5(19 + 2T− T^2)95)=0$

open the bracket and solve for T

$25+100T+9025+950T-475T^2=0\\475T^2-1050T-9050=0$

solving quadratic

T=5.6080

$distance\phantom{i}t=5.6080\\$$\\r = 5(1 + 4t)i + 5(19 + 2t − t^2)j\\r=r = 5(1 + 4(5.6080))i + 5(19 + 2(5.6080) − (5.6080)^2)j\\ r=117.16i-6.16832j\\distance=\sqrt{(117.16)^2+(-6.16832)^2}\\distance=\sqrt{13764.51377}\\distance=117.322$