Integrating vectors In a nutshell Calculus can be used with vectors to solve problems in 2D with variable acceleration. The general form of a vector is f ( t ) i + g ( t ) j f(t) \bold i + g(t)\bold j f ( t ) i + g ( t ) j
Equations DESCRIPTION EQUATION Displacement vector from the velocity vector.
s = ∫ v d t \bold s = \int \bold v \ dt s = ∫ v d t
Velocity vector from the acceleration vector.
v = ∫ a d t \bold v = \int \bold a \ dt v = ∫ a d t
Integrating vectors Note: Remember that the constant of integration must be obtained from information given. It will be a vector, so you should write it in vector form: K = p i + q j \bold K = p\bold i + q \bold j K = p i + q j
Example 1 The velocity vector of a body is v = ( ( 10 t 2 − 5 ) i + 7 j ) m s − 1 \bold v = ((10t^2 - 5 )\bold i + 7 \bold j) \ ms^{-1} v = (( 10 t 2 − 5 ) i + 7 j ) m s − 1 t = 0 s t=0 \ s t = 0 s r = ( 5 i − j ) m \bold r = (5\bold i - \bold j ) \ m r = ( 5 i − j ) m t . t. t .
Use r = ∫ v d t \bold r = \int \bold v \ dt r = ∫ v d t
r = ∫ v d t = ∫ ( ( 10 t 2 − 5 ) i + 7 j ) d t = ( 10 t 3 3 − 5 t ) i + 7 t j + K = ( 10 t 3 3 − 5 t + p ) i + ( 7 t + q ) j \begin{aligned}\bold r &= \int \bold v \ dt\\& = \int ((10t^2 - 5)\bold i + 7 \bold j) \ dt \\&=\left( \cfrac{10 t^3}{3} -5t\right)\bold i + 7t\bold j + \bold K \\&= \left( \cfrac{10 t^3}{3} -5t + p\right)\bold i + (7t + q)\bold j \end{aligned} r = ∫ v d t = ∫ (( 10 t 2 − 5 ) i + 7 j ) d t = ( 3 10 t 3 − 5 t ) i + 7 t j + K = ( 3 10 t 3 − 5 t + p ) i + ( 7 t + q ) j
Given the inital conditions that r = ( 5 i − j ) m \bold r = (5\bold i - \bold j ) \ m r = ( 5 i − j ) m t = 0 s t=0\ s t = 0 s
r ( t = 0 s ) = 5 i − j = ( 10 × 0 3 3 − 5 × 0 + p ) i + ( 7 × 0 + q ) j \begin{aligned}\bold r (t=0\ s ) &= 5\bold i - \bold j \\&= \left( \cfrac{10 \times 0^3}{3} -5\times 0 + p\right)\bold i + (7\times 0+ q)\bold j \end{aligned} r ( t = 0 s ) = 5 i − j = ( 3 10 × 0 3 − 5 × 0 + p ) i + ( 7 × 0 + q ) j
Therefore:
{ 5 i = p i − j = q j } ⟹ { p = 5 q = − 1 \begin{Bmatrix}\begin{aligned}5\bold i &= p \bold i \\-\bold j &= q\bold j \end{aligned}\end{Bmatrix}\implies \begin{cases}p = 5 \\q = -1\end{cases} { 5 i − j = p i = q j } ⟹ { p = 5 q = − 1
Thus, the position vector is:
r = [ ( 10 t 3 3 − 5 t + 5 ) i + ( 7 t − 1 ) j ] m ‾ \underline{\bold r = \left[ \left(\cfrac{10 t^3}{3} - 5t + 5 \right) \bold i + (7t-1)\bold j\right] \ m} r = [ ( 3 10 t 3 − 5 t + 5 ) i + ( 7 t − 1 ) j ] m
Example 2 The acceleration vector of a given body is a = [ ( 4 t − 2 ) i + 8 cos ( t ) j ] m s − 2 \bold a =\left[ (4t-2)\bold i + 8 \cos (t)\ \bold j\right] \ ms^{-2} a = [ ( 4 t − 2 ) i + 8 cos ( t ) j ] m s − 2 t = π t=\pi t = π
i) Calculate the velocity vector.
ii) Calculate the speed of this body when t = π 2 s t = \cfrac{\pi}{2} \ s t = 2 π s
i) Calculate the velocity vector.
Knowing that v = ∫ a d t \bold v = \int \bold a \ dt v = ∫ a d t
v = ∫ a d t = ∫ ( 4 t − 2 ) i + 8 cos ( t ) j d t = ( 4 t 2 2 − 2 t ) i + 8 sin ( t ) j + K = [ ( 2 t 2 − 2 t + p ) i + ( 8 sin t + q ) j ] m s − 1 \begin{aligned}\bold v &= \int \bold a \ dt \\&= \int (4t-2)\ \bold i + 8 \cos (t)\ \bold j \ dt \\&=\left(\cfrac{4t^2}{2} - 2t\right)\bold i + 8 \sin( t) \ \bold j + \bold K \\& = \left[\left(2t^2 - 2t+p\right )\bold i+ (8 \sin t+q)\ \bold j\right] \ ms^{-1}\end{aligned} v = ∫ a d t = ∫ ( 4 t − 2 ) i + 8 cos ( t ) j d t = ( 2 4 t 2 − 2 t ) i + 8 sin ( t ) j + K = [ ( 2 t 2 − 2 t + p ) i + ( 8 sin t + q ) j ] m s − 1
It is given that the body is at rest when t = π s . t = \pi \ s. t = π s .
v ( t = π s ) = 0 i + 0 j = = ( 2 π 2 − 2 π + p ) i + ( 8 sin π + q ) \begin{aligned}\bold v (t=\pi\ s) &= 0\bold i + 0 \bold j = \\&= (2\pi^2 - 2\pi + p)\bold i + (8\sin \pi + q)\end{aligned} v ( t = π s ) = 0 i + 0 j = = ( 2 π 2 − 2 π + p ) i + ( 8 sin π + q )
Which means:
{ 0 i = ( 2 π 2 − 2 π + p ) i 0 j = ( 8 sin π + q ) j } ⟹ { p = 2 π ( 1 − π ) q = 0 \begin{Bmatrix}\begin{aligned}0 \bold i & = (2\pi ^2 -2\pi + p)\bold i\\0 \bold j & = (\cancel{8\sin \pi} + q )\bold j\end{aligned}\end{Bmatrix}\implies\begin{cases}p = 2\pi (1-\pi)\\q = 0\end{cases} { 0 i 0 j = ( 2 π 2 − 2 π + p ) i = ( 8 sin π + q ) j } ⟹ { p = 2 π ( 1 − π ) q = 0
Thus:
v = [ ( 2 t 2 − 2 t + 2 π ( 1 − π ) ) i + 8 sin ( t ) j ] m s − 1 ‾ \underline{\bold v = \left[(2t^2 - 2t +2\pi (1-\pi))\bold i + 8\sin (t) \ \bold j\right]\ ms^{-1}} v = [ ( 2 t 2 − 2 t + 2 π ( 1 − π )) i + 8 sin ( t ) j ] m s − 1
ii) Calculate the speed of this body when t = π 2 s t = \cfrac{\pi}{2} \ s t = 2 π s
First, calculate the velocity at that time. Substitute t = π 2 t=\cfrac{\pi } {2} t = 2 π :
v ( t = π 2 ) = [ π ( − 3 π 2 + 1 ) i + 8 j ] m s − 1 \bold v\left(t = \cfrac{\pi}{2}\right) = \left[\pi\left(-\cfrac{3\pi}{2} + 1\right)\bold i + 8 \bold j\right]\ ms^{-1} v ( t = 2 π ) = [ π ( − 2 3 π + 1 ) i + 8 j ] m s − 1
Finally, find the speed using S p e e d = v x 2 + v y 2 Speed = \sqrt{v_x^2 + v_y^2} Sp ee d = v x 2 + v y 2
S p e e d = [ π ( − 3 π 2 + 1 ) ] 2 + 8 2 = 14.14 m s − 1 ( 2 d . p . ) ‾ Speed = \sqrt{\left[\pi\left(-\cfrac{3\pi}{2} + 1 \right) \right]^2 + 8^2} = \underline{14.14 \ ms^{-1} \ (2 \ d.p.)} Sp ee d = [ π ( − 2 3 π + 1 ) ] 2 + 8 2 = 14.14 m s − 1 ( 2 d . p . )
