Question

if position varies with time
as x = 4t^3+2t^2+2t+9
find the average . velocity from t=
2sec to t = 5sec.​

Answer 1

Answer:

$\boxed{\mathfrak{Average \ velocity \ (v_{avg}) = 172 \ m/s}}$

Explanation:

Position w.r.t. time:

$\rm x = 4 {t}^{3} + 2 {t}^{2} + 2t + 9$

$\rm At \: t_1 = 2s, \\ \rm x_1 = 4( {2}^{3} ) + 2( {2}^{2} ) + 2 \times 2 + 9 \\ \\ \rm = 4 \times 8 + 2 \times 4 + 4 + 9 \\ \\ \rm = 32 + 8 + 4 + 9 \\ \\ \rm = 53 \: m \\ \\ \rm At \: t_2 = 5s, \\ \rm x_2 = 4( {5}^{3} ) + 2( {5}^{2} ) + 2 \times 5 + 9 \\ \\ \rm = 4 \times 125 + 2 \times 25 + 10 + 9 \\ \\ \rm = 500 + 50 + 10 + 9 \\ \\ \rm = 569 \: m$

$\rm Displacement \: (\Delta x) = x_2 - x_1 \\ \\ \rm \Delta x= 569 - 53 \\ \\ \rm \Delta x= 516$

$\rm Time \: interval \: (\Delta t) = t_2 - t_1 \\ \\ \rm \Delta t = 5 - 2 \\ \\ \rm \Delta t =3s$

Average velocity is defined as the change in position or displacement of object divided by the time interval.

$\bf \implies v_{avg} = \dfrac{\Delta x}{\Delta t} \\ \\ \rm \implies v_{avg} = \dfrac{516}{3} \\ \\ \rm \implies v_{avg} = 172 \: m {s}^{ - 1}$

Answer 2

Average velocity = 172 m/s.

Explanation:

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Position varies with time :

↪ x = 4t³ + 2t² + 2t + 9

[ At t = 2sec ]

↪$x_1$ = 4(2)³ + 2(2)² + 2(2) + 9

↪$x_1$ = 4(8) + 2(4) + 4 + 9

↪$x_1$ = 32 + 8 + 13

↪$\tt\bold{x_1 = 59m}$

[ At t = 5sec ]

↪ $x_2$ = 4(5)³ + 2(5)² + 2(5) + 9

↪ $x_2$ = 4(125) + 2(25) + 10 + 9

↪ $x_2$ = 500 + 50 + 19

↪ $\tt\bold{x_2= 569m}$

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As we know that,

$\tt \bigstar \: \bold{Displacement (Δx) = x_2 - x_1}$

↪ Δx = 569 - 59

↪ $\tt\bold{Δx = 516m}$

$\tt \bigstar \bold{Time \: interval (Δt) = t_2 - t_1}$

↪ Δt = 5 - 2

↪ $\tt\bold{Δt = 3sec}$

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Now, we know that,

$\tt \: \bigstar \bold{Average \: velocity = \frac{Displacement(Δx)}{Time \: Interval(Δt) }}$

↪Average velocity = 516/3

$\large { \boxed{ \red {\frak{Average \: velocity = 172ms {}^{ - 1} }}}}$

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