Question #111630
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.
1
Expert's answer
2020-04-30T19:14:05-0400

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


r=ti+(8tt2)jr = ti + (8t − t^2)j

averageivelocity=r(3)r(2)32average\phantom{i}velocity=\frac{r(3)-r(2)}{3-2}\\

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


hence


averageivelocity=r(3)r(2)32averageivelocity=(3i+15j)(2i+12j)32averageivelocity=i+3j1averageivelocity=i+3javerage\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+(8tt2)jr = ti + (8t − t^2)j

velocity=drdtvelocity=drdt=i+(82t)jvelocity=\frac{dr}{dt}\\velocity=\frac{dr}{dt}=i+(8-2t)j


hence at t=4

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


for speed

speed=drdtdrdt=ispeed=|\frac{dr}{dt}|\\\frac{dr}{dt}=i

hence speed

speed=12    speed=1speed=\sqrt{1^2}\implies speed=1


direction of motion

direction=v(t)v(t)direction=i1direction=idirection=\frac{v(t)}{|v(t)|}\\direction=\frac{i}{1}\\direction=i

tanθ=ycomponentxcomponentiofitheivelocityθ=tan1(01)θ=00tan\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

(t8tt2)=k(12)    t=k8tt2=2k\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

8tt2=2thencet=10r=10i20j8t-t^2=-2t\\hence\\t=10 \\\therefore r=10i-20j

d.) acceleration vector

acceleration=dvdtdvdt=2jacceleration=\frac{dv}{dt}\\\frac{dv}{dt}=-2j



2.) initial velocity

r=5(1+4t)i+5(19+2tt2)j.velocity=drdtdrdt=20i+5(22t)jvelocity=20i+5(22t)jatit=0r = 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=20i10tjv=20i-10tj


value of T

r=5(1+4t)i+5(19+2tt2)jr = 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=0r=5i+95jt=0\\r = 5i + 95j

at t=T

r=5(1+4T)i+5(19+2TT2)jr = 5(1 + 4T)i + 5(19 + 2T − T^2)j

then

(5(1+4T)i+5(19+2tT2)j).(5i+95j)=0(5(1+4T)5)+(5(19+2TT2)95)=0(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+950T475T2=0475T21050T9050=025+100T+9025+950T-475T^2=0\\475T^2-1050T-9050=0

solving quadratic

T=5.6080

distanceit=5.6080distance\phantom{i}t=5.6080\\r=5(1+4t)i+5(19+2tt2)jr=r=5(1+4(5.6080))i+5(19+2(5.6080)(5.6080)2)jr=117.16i6.16832jdistance=(117.16)2+(6.16832)2distance=13764.51377distance=117.322\\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


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