(a) average velocity from t = 2 to t = 3,
r = t i + ( 8 t − t 2 ) j r = ti + (8t − t^2)j r = t i + ( 8 t − t 2 ) j
a v e r a g e i v e l o c i t y = r ( 3 ) − r ( 2 ) 3 − 2 average\phantom{i}velocity=\frac{r(3)-r(2)}{3-2}\\ a v er a g e i v e l oc i t y = 3 − 2 r ( 3 ) − r ( 2 )
r ( 3 ) = 3 i + ( 8 ( 3 ) − ( 3 ) 2 ) j r ( 3 ) = 3 i + 15 j r ( 2 ) = 2 i + ( 8 ( 2 ) − ( 2 ) 2 ) j r ( 2 ) = 2 i + 12 j r(3) = 3i + (8(3) − (3)^2)j\\r(3) = 3i + 15j\\r(2)= 2i + (8(2) −(2)^2)j\\r(2) = 2i + 12j r ( 3 ) = 3 i + ( 8 ( 3 ) − ( 3 ) 2 ) j r ( 3 ) = 3 i + 15 j r ( 2 ) = 2 i + ( 8 ( 2 ) − ( 2 ) 2 ) j r ( 2 ) = 2 i + 12 j
hence
a v e r a g e i v e l o c i t y = r ( 3 ) − r ( 2 ) 3 − 2 a v e r a g e i v e l o c i t y = ( 3 i + 15 j ) − ( 2 i + 12 j ) 3 − 2 a v e r a g e i v e l o c i t y = i + 3 j 1 a v e r a g e i v e l o c i t y = i + 3 j 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 a v er a g e i v e l oc i t y = 3 − 2 r ( 3 ) − r ( 2 ) a v er a g e i v e l oc i t y = 3 − 2 ( 3 i + 15 j ) − ( 2 i + 12 j ) a v er a g e i v e l oc i t y = 1 i + 3 j a v er a g e i v e l oc i t y = i + 3 j
(b) velocity, speed and direction of motion when t = 4,
velocity function
r = t i + ( 8 t − t 2 ) j r = ti + (8t − t^2)j r = t i + ( 8 t − t 2 ) j
v e l o c i t y = d r d t v e l o c i t y = d r d t = i + ( 8 − 2 t ) j velocity=\frac{dr}{dt}\\velocity=\frac{dr}{dt}=i+(8-2t)j v e l oc i t y = d t d r v e l oc i t y = d t d r = i + ( 8 − 2 t ) j
hence at t=4
v e l o c i t y = i + 0 j ⟹ v e l o c i t y = i velocity=i+0j\implies velocity=i v e l oc i t y = i + 0 j ⟹ v e l oc i t y = i
for speed
s p e e d = ∣ d r d t ∣ d r d t = i speed=|\frac{dr}{dt}|\\\frac{dr}{dt}=i s p ee d = ∣ d t d r ∣ d t d r = i
hence speed
s p e e d = 1 2 ⟹ s p e e d = 1 speed=\sqrt{1^2}\implies speed=1 s p ee d = 1 2 ⟹ s p ee d = 1
direction of motion
d i r e c t i o n = v ( t ) ∣ v ( t ) ∣ d i r e c t i o n = i 1 d i r e c t i o n = i direction=\frac{v(t)}{|v(t)|}\\direction=\frac{i}{1}\\direction=i d i rec t i o n = ∣ v ( t ) ∣ v ( t ) d i rec t i o n = 1 i d i rec t i o n = i
t a n θ = y − c o m p o n e n t x − c o m p o n e n t i o f i t h e i v e l o c i t y θ = t a n − 1 ( 0 1 ) θ = 0 0 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} t an θ = x − co m p o n e n t y − co m p o n e n t i o f i t h e i v e l oc i t y θ = t a n − 1 ( 1 0 ) θ = 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
( t 8 t − t 2 ) = k ( 1 − 2 ) ⟹ t = k 8 t − t 2 = − 2 k \begin{pmatrix}
t \\
8t-t^2
\end{pmatrix}=k\begin{pmatrix}
1 \\
-2
\end{pmatrix}\\\implies t=k\\8t-t^2=-2k\\ ( t 8 t − t 2 ) = k ( 1 − 2 ) ⟹ t = k 8 t − t 2 = − 2 k
solving the two equations simultaneously by substitution
8 t − t 2 = − 2 t h e n c e t = 10 ∴ r = 10 i − 20 j 8t-t^2=-2t\\hence\\t=10
\\\therefore r=10i-20j 8 t − t 2 = − 2 t h e n ce t = 10 ∴ r = 10 i − 20 j
d.) acceleration vector
a c c e l e r a t i o n = d v d t d v d t = − 2 j acceleration=\frac{dv}{dt}\\\frac{dv}{dt}=-2j a cce l er a t i o n = d t d v d t d v = − 2 j
2.) initial velocity
r = 5 ( 1 + 4 t ) i + 5 ( 19 + 2 t − t 2 ) j . v e l o c i t y = d r d t d r d t = 20 i + 5 ( 2 − 2 t ) j v e l o c i t y = 20 i + 5 ( 2 − 2 t ) j a t i t = 0 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 r = 5 ( 1 + 4 t ) i + 5 ( 19 + 2 t − t 2 ) j . v e l oc i t y = d t d r d t d r = 20 i + 5 ( 2 − 2 t ) j v e l oc i t y = 20 i + 5 ( 2 − 2 t ) j a t i t = 0
initial velocity
v = 20 i − 10 t j v=20i-10tj v = 20 i − 10 t j
value of T
r = 5 ( 1 + 4 t ) i + 5 ( 19 + 2 t − t 2 ) j r = 5(1 + 4t)i + 5(19 + 2t − t^2)j r = 5 ( 1 + 4 t ) i + 5 ( 19 + 2 t − 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 = 5 i + 95 j t=0\\r = 5i + 95j t = 0 r = 5 i + 95 j
at t=T
r = 5 ( 1 + 4 T ) i + 5 ( 19 + 2 T − T 2 ) j r = 5(1 + 4T)i + 5(19 + 2T − T^2)j r = 5 ( 1 + 4 T ) i + 5 ( 19 + 2 T − T 2 ) j
then
( 5 ( 1 + 4 T ) i + 5 ( 19 + 2 t − T 2 ) j ) . ( 5 i + 95 j ) = 0 ( 5 ( 1 + 4 T ) 5 ) + ( 5 ( 19 + 2 T − T 2 ) 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 ( 5 ( 1 + 4 T ) i + 5 ( 19 + 2 t − T 2 ) j ) . ( 5 i + 95 j ) = 0 ( 5 ( 1 + 4 T ) 5 ) + ( 5 ( 19 + 2 T − T 2 ) 95 ) = 0
open the bracket and solve for T
25 + 100 T + 9025 + 950 T − 475 T 2 = 0 475 T 2 − 1050 T − 9050 = 0 25+100T+9025+950T-475T^2=0\\475T^2-1050T-9050=0 25 + 100 T + 9025 + 950 T − 475 T 2 = 0 475 T 2 − 1050 T − 9050 = 0
solving quadratic
T=5.6080
d i s t a n c e i t = 5.6080 distance\phantom{i}t=5.6080\\ d i s t an ce i t = 5.6080 r = 5 ( 1 + 4 t ) i + 5 ( 19 + 2 t − t 2 ) j r = r = 5 ( 1 + 4 ( 5.6080 ) ) i + 5 ( 19 + 2 ( 5.6080 ) − ( 5.6080 ) 2 ) j r = 117.16 i − 6.16832 j d i s t a n c e = ( 117.16 ) 2 + ( − 6.16832 ) 2 d i s t a n c e = 13764.51377 d i s t a n c e = 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 r = 5 ( 1 + 4 t ) i + 5 ( 19 + 2 t − t 2 ) j r = r = 5 ( 1 + 4 ( 5.6080 )) i + 5 ( 19 + 2 ( 5.6080 ) − ( 5.6080 ) 2 ) j r = 117.16 i − 6.16832 j d i s t an ce = ( 117.16 ) 2 + ( − 6.16832 ) 2 d i s t an ce = 13764.51377 d i s t an ce = 117.322
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