In electric and autonomous vehicle engineering, onboard inertial measurement units (IMUs) log high-frequency 3-axis accelerometer data. When GPS satellite signals are occluded (such as inside tunnels or subterranean parking garages), vehicle stability control systems reconstruct the vehicle's instantaneous velocity and position trajectory by performing successive vector integrations over time (dead reckoning).
An autonomous electric delivery vehicle enters an underground facility at t = 0 Β seconds t = 0\text{ seconds} t = 0 Β seconds m/s 2 \text{m/s}^2 m/s 2
a β ( t ) = β¨ 6 t , 12 t 2 , 4 β© \vec{a}(t) = \langle 6t, 12t^2, 4 \rangle a ( t ) = β¨ 6 t , 12 t 2 , 4 β©
At the moment of entering the facility (
t = 0 t = 0 t = 0 ), the vehicle's initial velocity is
v β ( 0 ) = β¨ 2 , β 1 , 5 β© Β m/s \vec{v}(0) = \langle 2, -1, 5 \rangle\text{ m/s} v ( 0 ) = β¨ 2 , β 1 , 5 β© Β m/s and its initial position vector relative to the entrance gate is
r β ( 0 ) = β¨ 0 , 10 , β 3 β© Β meters \vec{r}(0) = \langle 0, 10, -3 \rangle\text{ meters} r ( 0 ) = β¨ 0 , 10 , β 3 β© Β meters .
Use indefinite vector integration and the initial velocity condition to derive the vehicle's 3D velocity vector function v β ( t ) \vec{v}(t) v ( t )
Use indefinite vector integration and the initial position condition to derive the vehicle's 3D position trajectory function r β ( t ) \vec{r}(t) r ( t )
Calculate the net displacement vector Ξ r β \Delta \vec{r} Ξ r 0 β€ t β€ 3 0 \le t \le 3 0 β€ t β€ 3
Step-by-Step Solution:
Derive Velocity Vector Function v β ( t ) \vec{v}(t) v ( t )
Integrate acceleration component-by-component:
v β ( t ) = β« a β ( t ) β d t = β¨ β« 6 t β d t , β« 12 t 2 β d t , β« 4 β d t β© = β¨ 3 t 2 , 4 t 3 , 4 t β© + C β 1 \vec{v}(t) = \int \vec{a}(t) \, dt = \left\langle \int 6t \, dt, \int 12t^2 \, dt, \int 4 \, dt \right\rangle = \langle 3t^2, 4t^3, 4t \rangle + \vec{C}_1 v ( t ) = β« a ( t ) d t = β¨ β« 6 t d t , β« 12 t 2 d t , β« 4 d t β© = β¨ 3 t 2 , 4 t 3 , 4 t β© + C 1 β v β ( 0 ) = β¨ 2 , β 1 , 5 β© \vec{v}(0) = \langle 2, -1, 5 \rangle v ( 0 ) = β¨ 2 , β 1 , 5 β© v β ( 0 ) = β¨ 0 , 0 , 0 β© + C β 1 = β¨ 2 , β 1 , 5 β© β
β βΉ β
β C β 1 = β¨ 2 , β 1 , 5 β© \vec{v}(0) = \langle 0, 0, 0 \rangle + \vec{C}_1 = \langle 2, -1, 5 \rangle \implies \vec{C}_1 = \langle 2, -1, 5 \rangle v ( 0 ) = β¨ 0 , 0 , 0 β© + C 1 β = β¨ 2 , β 1 , 5 β© βΉ C 1 β = β¨ 2 , β 1 , 5 β© v β ( t ) = β¨ 3 t 2 + 2 , 4 t 3 β 1 , 4 t + 5 β© Β m/s \vec{v}(t) = \langle 3t^2 + 2, 4t^3 - 1, 4t + 5 \rangle \text{ m/s} v ( t ) = β¨ 3 t 2 + 2 , 4 t 3 β 1 , 4 t + 5 β© Β m/s Derive Position Vector Function r β ( t ) \vec{r}(t) r ( t )
Integrate velocity component-by-component:
r β ( t ) = β« v β ( t ) β d t = β¨ β« ( 3 t 2 + 2 ) β d t , β« ( 4 t 3 β 1 ) β d t , β« ( 4 t + 5 ) β d t β© = β¨ t 3 + 2 t , t 4 β t , 2 t 2 + 5 t β© + C β 2 \vec{r}(t) = \int \vec{v}(t) \, dt = \left\langle \int (3t^2 + 2) \, dt, \int (4t^3 - 1) \, dt, \int (4t + 5) \, dt \right\rangle = \langle t^3 + 2t, t^4 - t, 2t^2 + 5t \rangle + \vec{C}_2 r ( t ) = β« v ( t ) d t = β¨ β« ( 3 t 2 + 2 ) d t , β« ( 4 t 3 β 1 ) d t , β« ( 4 t + 5 ) d t β© = β¨ t 3 + 2 t , t 4 β t , 2 t 2 + 5 t β© + C 2 β r β ( 0 ) = β¨ 0 , 10 , β 3 β© \vec{r}(0) = \langle 0, 10, -3 \rangle r ( 0 ) = β¨ 0 , 10 , β 3 β© r β ( 0 ) = β¨ 0 , 0 , 0 β© + C β 2 = β¨ 0 , 10 , β 3 β© β
β βΉ β
β C β 2 = β¨ 0 , 10 , β 3 β© \vec{r}(0) = \langle 0, 0, 0 \rangle + \vec{C}_2 = \langle 0, 10, -3 \rangle \implies \vec{C}_2 = \langle 0, 10, -3 \rangle r ( 0 ) = β¨ 0 , 0 , 0 β© + C 2 β = β¨ 0 , 10 , β 3 β© βΉ C 2 β = β¨ 0 , 10 , β 3 β© r β ( t ) = β¨ t 3 + 2 t , t 4 β t + 10 , 2 t 2 + 5 t β 3 β© Β meters \vec{r}(t) = \langle t^3 + 2t, t^4 - t + 10, 2t^2 + 5t - 3 \rangle \text{ meters} r ( t ) = β¨ t 3 + 2 t , t 4 β t + 10 , 2 t 2 + 5 t β 3 β© Β meters Calculate Net Displacement Ξ r β \Delta \vec{r} Ξ r [ 0 , 3 ] [0, 3] [ 0 , 3 ]
Evaluate the definite integral of velocity:
Ξ r β = β« 0 3 v β ( t ) β d t = r β ( 3 ) β r β ( 0 ) \Delta \vec{r} = \int_{0}^{3} \vec{v}(t) \, dt = \vec{r}(3) - \vec{r}(0) Ξ r = β« 0 3 β v ( t ) d t = r ( 3 ) β r ( 0 ) r β ( 3 ) \vec{r}(3) r ( 3 )
x ( 3 ) = ( 3 ) 3 + 2 ( 3 ) = 27 + 6 = 33 x(3) = (3)^3 + 2(3) = 27 + 6 = 33 x ( 3 ) = ( 3 ) 3 + 2 ( 3 ) = 27 + 6 = 33 y ( 3 ) = ( 3 ) 4 β ( 3 ) + 10 = 81 β 3 + 10 = 88 y(3) = (3)^4 - (3) + 10 = 81 - 3 + 10 = 88 y ( 3 ) = ( 3 ) 4 β ( 3 ) + 10 = 81 β 3 + 10 = 88 z ( 3 ) = 2 ( 3 ) 2 + 5 ( 3 ) β 3 = 18 + 15 β 3 = 30 z(3) = 2(3)^2 + 5(3) - 3 = 18 + 15 - 3 = 30 z ( 3 ) = 2 ( 3 ) 2 + 5 ( 3 ) β 3 = 18 + 15 β 3 = 30 r β ( 3 ) = β¨ 33 , 88 , 30 β© Β m \vec{r}(3) = \langle 33, 88, 30 \rangle \text{ m} r ( 3 ) = β¨ 33 , 88 , 30 β© Β m r β ( 0 ) = β¨ 0 , 10 , β 3 β© \vec{r}(0) = \langle 0, 10, -3 \rangle r ( 0 ) = β¨ 0 , 10 , β 3 β© Ξ r β = β¨ 33 β 0 , 88 β 10 , 30 β ( β 3 ) β© = β¨ 33 , 78 , 33 β© Β meters \Delta \vec{r} = \langle 33 - 0, 88 - 10, 30 - (-3) \rangle = \langle 33, 78, 33 \rangle \text{ meters} Ξ r = β¨ 33 β 0 , 88 β 10 , 30 β ( β 3 )β© = β¨ 33 , 78 , 33 β© Β meters
Meaning of the Answer & Real-Life Application:
Over the 3-second interval, the vehicle undergoes a 3D spatial translation of β¨ 33 , 78 , 33 β© Β meters \langle 33, 78, 33 \rangle\text{ meters} β¨ 33 , 78 , 33 β© Β meters β₯ Ξ r β β₯ = 33 2 + 78 2 + 33 2 = 1089 + 6084 + 1089 = 8262 β 90.9 Β meters \|\Delta \vec{r}\| = \sqrt{33^2 + 78^2 + 33^2} = \sqrt{1089 + 6084 + 1089} = \sqrt{8262} \approx 90.9\text{ meters} β₯Ξ r β₯ = 3 3 2 + 7 8 2 + 3 3 2 β = 1089 + 6084 + 1089 β = 8262 β β 90.9 Β meters