Speed of the Plane
Given,
The model for the path of the plane during this maneuver is
r ( t ) = 10 c o s 10 π i ⃗ + 10 s i n 10 π t j ⃗ + ( 4 + 4 t ) k ⃗ , 0 ≤ t ≤ 1 20 r(t) = 10 cos 10\pi \space \vec i + 10 sin 10\pi t \space \vec j + (4 + 4t ) \vec k , \space 0\le t\le\frac {1}{20} r ( t ) = 10 cos 10 π i + 10 s in 10 π t j + ( 4 + 4 t ) k , 0 ≤ t ≤ 20 1 Here t is the time in hours and
r is the position vector
(a)
We can calculate the speed of the plane by the formula
∣ ∣ r ′ ( t ) ∣ ∣ ||r' (t)|| ∣∣ r ′ ( t ) ∣∣
So, Differentiate the position vector r(t) with respect to t
v ( t ) = r ′ ( t ) = d d t ( 10 c o s 10 π i ⃗ + 10 s i n 10 π t j ⃗ + ( 4 + 4 t ) k ⃗ ) v(t) = r' (t) = \frac {d}{dt} (10 cos 10\pi \space \vec i + 10 sin 10\pi t \space \vec j + (4 + 4t ) \vec k ) v ( t ) = r ′ ( t ) = d t d ( 10 cos 10 π i + 10 s in 10 π t j + ( 4 + 4 t ) k )
= − 100 π s i n 10 π t i ⃗ + 100 π c o s 10 π t j ⃗ + 4 k ⃗ = - 100\pi sin 10\pi \space t \space \vec i \space + \space 100 \pi cos 10\pi t \space \vec j + 4 \vec k = − 100 π s in 10 π t i + 100 π cos 10 π t j + 4 k
Now the Speed of the plane =
∣ ∣ v ( t ) ∣ ∣ = ∣ ∣ r ′ ( t ) ∣ ∣ = ∣ ∣ − 100 π s i n 10 π t i ⃗ + 100 π c o s 10 π t j ⃗ + 4 k ⃗ ∣ ∣ ||v(t) || = ||r' (t) || = || - 100 \pi sin 10\pi \space t \space \vec i \space + \space 100\pi cos 10\pi t \space \vec j + 4 \vec k \space || ∣∣ v ( t ) ∣∣ = ∣∣ r ′ ( t ) ∣∣ = ∣∣ − 100 π s in 10 π t i + 100 π cos 10 π t j + 4 k ∣∣
= ( − 100 π s i n 10 π t ) 2 + ( 100 π c o s 10 π t ) 2 + 4 2 = \sqrt {(- 100 \pi sin10\pi t)^2 + (100\pi cos 10\pi t)^2 + 4^2 } = ( − 100 π s in 10 π t ) 2 + ( 100 π cos 10 π t ) 2 + 4 2 = ( 10000 π 2 s i n 2 10 π t ) + ( 10000 π 2 c o s 2 10 π t ) + 16 = \sqrt {( 10000 \pi^2 sin^2 10\pi t) + (10000\pi^2 cos^2 10\pi t) + 16 } = ( 10000 π 2 s i n 2 10 π t ) + ( 10000 π 2 co s 2 10 π t ) + 16
= 10000 π 2 [ ( s i n 2 ( 10 π t ) + c o s 2 ( 10 π t ) ] + 16 = \sqrt {10000 \pi^2 \space [( sin^2 (10\pi t ) + cos^2 (10\pi t)] + 16} = 10000 π 2 [( s i n 2 ( 10 π t ) + co s 2 ( 10 π t )] + 16
∣ ∣ v ( t ) ∣ ∣ = 10000 π 2 ( 1 ) + 16 = 10000 π 2 + 16 ||v(t) || = \sqrt {10000\pi^2 (1) + 16} = \sqrt {10000\pi^2 + 16} ∣∣ v ( t ) ∣∣ = 10000 π 2 ( 1 ) + 16 = 10000 π 2 + 16
N o w t h e S p e e d o f t h e p l a n e = = 4 × 625 π 2 + 1 = 314.025 m i l e s / h r Now \space the \space Speed \space of \space the \space plane = = 4 \times \sqrt {625\pi^2 + 1} = 314.025 \space miles/hr N o w t h e Sp ee d o f t h e pl an e == 4 × 625 π 2 + 1 = 314.025 mi l es / h r
(b).
Now, we need to calculate the α T a n d α N α_{T} \space and \space α_{N} α T an d α N
Here, formula for Tangential acceleration is given by
α T = d d t [ ∣ ∣ v ∣ ∣ ] = v . α ∣ ∣ v ∣ ∣ α_{T} = \frac {d}{dt} [||v||] = \frac {v.α}{||v||} α T = d t d [ ∣∣ v ∣∣ ] = ∣∣ v ∣∣ v . α
We know,
α ( t ) = v ′ ( t ) = d d t ( − 100 π s i n 10 π t i ⃗ + 100 π c o s 10 π t j ⃗ + 4 k ⃗ ) α(t) = v'(t) = \frac {d}{dt} (- 100 \pi sin 10\pi \space t \space \vec i \space + \space 100 \pi cos 10\pi t \space \vec j + 4 \vec k ) α ( t ) = v ′ ( t ) = d t d ( − 100 π s in 10 π t i + 100 π cos 10 π t j + 4 k )
= − 1000 π 2 c o s 10 π t i ⃗ − 1000 π 2 s i n 10 π t j ⃗ = -1000 \pi^2 \space cos 10\pi t \vec i - 1000 \pi^2 sin 10\pi t \space \vec j = − 1000 π 2 cos 10 π t i − 1000 π 2 s in 10 π t j
v . α = ( − 100 π s i n 10 π t i ⃗ + 100 π c o s 10 π t j ⃗ + 4 k ⃗ ) . ( − 1000 π 2 c o s 10 π t i ⃗ − 1000 π 2 s i n 10 π t j ⃗ ) v.α = (- 100\pi sin 10\pi \space t \space \vec i \space + \space 100 \pi cos 10\pi t \space \vec j + 4 \vec k ).
(-1000 \pi^2 \space cos 10\pi t \vec i - 1000 \pi^2 sin 10\pi t \space \vec j) v . α = ( − 100 π s in 10 π t i + 100 π cos 10 π t j + 4 k ) . ( − 1000 π 2 cos 10 π t i − 1000 π 2 s in 10 π t j )
= 100000 π 3 s i n ( 10 π t ) c o s ( 10 π t ) − 100000 π 3 s i n ( 10 π t ) c o s ( 10 π t ) + 0 = 100000\pi^3 sin (10\pi t) \space cos (10\pi t) - 100000 \pi^3 sin (10\pi t) cos (10\pi t) + 0 = 100000 π 3 s in ( 10 π t ) cos ( 10 π t ) − 100000 π 3 s in ( 10 π t ) cos ( 10 π t ) + 0
v . α = 0 v.α = 0 v . α = 0
So, Tangential acceleration is
α T = v . α ∣ ∣ v ∣ ∣ = 0 314.025 = 0 α_{T} = \frac {v.α}{||v||} = \frac {0} { 314.025} = 0 α T = ∣∣ v ∣∣ v . α = 314.025 0 = 0
Here, formula for Normal components of acceleration is given by
α N = ∣ ∣ v × α ∣ ∣ ∣ ∣ v ∣ ∣ α_{N} = \frac {||v \times α||}{||v||} α N = ∣∣ v ∣∣ ∣∣ v × α ∣∣
v × α = ∣ i j k − 100 π s i n ( 10 π t ) 100 π c o s ( 10 π t ) 4 − 1000 π 2 c o s ( 10 π t ) − 1000 π 2 s i n ( 10 π t ) 0 ∣ v \times α = \begin{vmatrix}
i & j & k \\
-100 \pi sin (10\pi t) & 100\pi cos (10\pi t) & 4 \\
-1000\pi^2 cos (10\pi t) & -1000 \pi^2 sin (10\pi t) & 0
\end{vmatrix} v × α = ∣ ∣ i − 100 π s in ( 10 π t ) − 1000 π 2 cos ( 10 π t ) j 100 π cos ( 10 π t ) − 1000 π 2 s in ( 10 π t ) k 4 0 ∣ ∣
= i ( 0 + 4000 π 2 s i n ( 10 π t ) ) − j ( 0 + 4000 π 2 c o s ( 10 π t ) + k ( 100000 π 3 s i n 2 ( 10 π t + 100000 π 3 c o s 2 ( 10 π t ) ) = i (0 + 4000 \pi^2 sin (10\pi t)) - j (0 + 4000\pi^2 cos (10\pi t) \\+ k ( 100000 \pi^3 sin^2 (10\pi t + 100000 \pi^3 cos^2 (10\pi t)) = i ( 0 + 4000 π 2 s in ( 10 π t )) − j ( 0 + 4000 π 2 cos ( 10 π t ) + k ( 100000 π 3 s i n 2 ( 10 π t + 100000 π 3 co s 2 ( 10 π t )) .
v × α = = i ( 4000 π 2 s i n ( 10 π t ) ) − j ( 4000 π 2 c o s ( 10 π t ) + k × 100000 π 3 ( s i n 2 ( 10 π t + c o s 2 ( 10 π t ) ) ) v \times α = = i ( 4000 \pi^2 sin (10\pi t)) - j ( 4000\pi^2 cos (10\pi t) \\+ k \times 100000 \pi^3 (sin^2 (10\pi t + cos^2 (10\pi t))) v × α == i ( 4000 π 2 s in ( 10 π t )) − j ( 4000 π 2 cos ( 10 π t ) + k × 100000 π 3 ( s i n 2 ( 10 π t + co s 2 ( 10 π t )))
v × α = i ( 4000 π 2 s i n ( 10 π t ) ) − j ( 4000 π 2 c o s ( 10 π t ) + k × 100000 π 3 v \times α = i ( 4000 \pi^2 sin (10\pi t)) - j ( 4000\pi^2 cos (10\pi t) \\+ k \times 100000 \pi^3 v × α = i ( 4000 π 2 s in ( 10 π t )) − j ( 4000 π 2 cos ( 10 π t ) + k × 100000 π 3
∣ ∣ v × α ∣ ∣ = ( 4000 π 2 ) 2 s i n 2 ( 10 π t ) + ( 4000 π 2 ) 2 c o s 2 ( 10 π t ) + 1000 0 2 π 6 || v \times α ||= \sqrt {(4000 \pi^2)^2 sin ^2(10\pi t) + (4000 \pi^2)^2 cos^2 (10\pi t) + 10000^2 \pi^6} ∣∣ v × α ∣∣ = ( 4000 π 2 ) 2 s i n 2 ( 10 π t ) + ( 4000 π 2 ) 2 co s 2 ( 10 π t ) + 1000 0 2 π 6
∣ ∣ v × α ∣ ∣ = ( 4000 π 2 ) 2 [ s i n 2 ( 10 π t ) + c o s 2 ( 10 π t ) ] + 1000 0 2 π 6 || v \times α ||= \sqrt {(4000 \pi^2)^2 \space [sin ^2(10\pi t) + cos^2 (10\pi t)] + 10000^2 \pi^6} ∣∣ v × α ∣∣ = ( 4000 π 2 ) 2 [ s i n 2 ( 10 π t ) + co s 2 ( 10 π t )] + 1000 0 2 π 6
∣ ∣ v × α ∣ ∣ = ( 4000 π 2 ) 2 + 1000 0 2 π 6 || v \times α ||= \sqrt {(4000 \pi^2)^2 \space + 10000^2 \pi^6} ∣∣ v × α ∣∣ = ( 4000 π 2 ) 2 + 1000 0 2 π 6
∣ ∣ v × α ∣ ∣ = 4000 π 2 1 + 625 π 2 || v \times α ||= 4000 \pi^2 \sqrt {1 +625 \pi^2} ∣∣ v × α ∣∣ = 4000 π 2 1 + 625 π 2
α N = ∣ ∣ v × α ∣ ∣ ∣ ∣ v ∣ ∣ = 4000 π 2 1 + 625 π 2 4 × 625 π 2 + 1 α_{N} = \frac {||v \times α||}{||v||} = \frac {4000\pi^2 \sqrt {1+ 625\pi^2}}{4 \times \sqrt {625\pi^2 + 1}} α N = ∣∣ v ∣∣ ∣∣ v × α ∣∣ = 4 × 625 π 2 + 1 4000 π 2 1 + 625 π 2
Normal components of acceleration is
α N = 1000 π 2 α_{N} = 1000 \pi^2 α N = 1000 π 2
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