Question

The displacement (in metre) of a particle moving along x-axis is given by x = 18t + 5t2 : calculate (i) the instantaneous velocity at t=2 sec (ii) average velocity between t=2 sec and t=3sec. (iii) instantaneous acceleration.

Answer 1

GiveN :

• x = 18t + 5t²

To FinD :

• Velocity at t = 2s
• Average velocity between t = 2s and t = 3s
• Instantenous acceleration

SolutioN :

Take given equation :

$\implies \rm{x = 18t + 5t^2}$

Differentiate wrt. dt

$\\ \implies \rm{\dfrac{dx}{dt} = \dfrac{d(18t + 5t^2)}{dt}} \\ \\ \\ \implies \rm{v = 2(5t) + 18} \\ \\ \\ \implies \rm{v \: = \: (10t \: + \: 18) \: ms^{-1} \: \: \: \: \: \: \: \: ...(1)}$

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Put t = 2s in (1)

$\implies \rm{v_1 = 10(2) + 18} \\ \\ \\ \implies \rm{v_1 = 20 + 18} \\ \\ \\ \implies \rm{v_1 \: = 38 \: ms^{-1}}$

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Put t = 3s in (1)

$\implies \rm{v_2 = 10(3) + 18} \\ \\ \\ \implies \rm{v_2 = 30 + 18} \\ \\ \\ \implies \rm{v_2 \: = \: 48 \: ms^{-1}}$

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Now, for finding average velocity between t = 2s and t = 3s

$\implies \rm{v_{avg} = \dfrac{v_1 + v_2}{2}} \\ \\ \\ \implies \rm{v_{avg} = \dfrac{48 + 38}{2}} \\ \\ \\ \implies \rm{v_{avg} = \dfrac{86}{2}} \\ \\ \\ \implies \rm{v_{avg} \: = \: 43 \: ms^{-1}}$

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Now, again differentiate (1) wrt. dt

$\implies \rm{\dfrac{dv}{dt} = \dfrac{d(10t + 18)}{dt}} \\ \\ \\ \implies \rm{a = 10 + 0} \\ \\ \\ \implies \rm{a \: = \: 10 \: ms^{-2}}$

Answer 2

GivEn:-

• The displacement (in metre) of a particle moving along x-axis is given by x = 18t + 5t²

To find:-

• the instantaneous velocity at t = 2 sec

• average velocity between t = 2 sec and t = 3sec.

• instantaneous acceleration.

SoluTion:-

GivEn that,

⛥ The displacement (in metre) of a particle moving along x-axis is given by x = 18t + 5t²

★ The givEn equation is a relation b/w distance and time.

For finding Relation b/w velocity and time, we will differentiate x w.r.t. t.

$:\implies\sf v = \dfrac{dx}{dt}$

$:\implies\sf v = \dfrac{d(18t + 5t^2)}{dt}$

$:\implies\sf v = 2(5t) + 18$

$:\implies\bf 10t + 18$

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Therefore,

✦ velocity at t = 2 sec is,

$:\implies\sf v_1 = 10 \times 2 + 18$

$:\implies\sf v_1 = 20 + 18$

$:\implies\sf v_1 = 20 + 18$

$:\implies{\underline{\boxed{\bf{\pink{v_1 = 38ms^{-1}}}}}}$ ★

Now,

✦velocity at t = 3 sec is,

$:\implies\sf v_2 = 10 \times 3 + 18$

$:\implies\sf v_2 = 30 + 18$

$:\implies\sf v_2 = 30 + 18$

$:\implies{\underline{\boxed{\bf{\pink{v_2 = 48ms^{-1}}}}}}$ ★

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Now,

✦ We have to find average velocity b/w t = 2sec and t = 3sec,

$:\implies\sf v_{average} = \dfrac{v_1 + v_2 }{2}$

$:\implies\sf v_{average} = \dfrac{38 + 48}{2}$

$:\implies\sf v_{average} = \dfrac{86}{2}$

$:\implies\sf v_{average} = \cancel{ \dfrac{86}{2}}$

$:\implies{\underline{\boxed{\bf{\blue{v_{average} = 43\; ms^{-1}}}}}}$ ★

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✦ Now again, we have to differentiate v = 10t + 18 w.r.t. dt,

$:\implies\sf \dfrac{dv}{dt} = \dfrac{d(10t + 18)}{dt}$

$:\implies\sf a = 10 + 0$

$:\implies{\underline{\boxed{\bf{\purple{a = 10ms^{-2}}}}}}$ ★

$\dag$ Hence SolvEd!!

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