Question

the displacement of a particle moving along x-axis is given by x=18t+5^2 the average acceleration during the interval t1=2sec. and t2=4sec is

Answer 1

Correct Question -

The displacement of a particle moving along x-axis is given by -

x = 18t + 5t² .

The average acceleration during the interval t¹ = 2 seconds and

t² = 4 seconds is -

In the above Question , the following information is given -

The displacement of a particle moving along x-axis is given by -

x = 18t + 5t²

To find -

The average acceleration during the interval t¹ = 2 seconds and

t² = 4 seconds .

Solution -

In the above Question , the equation for displacement is given as -

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

Now,

Differenciating displacement with respect to find gives velocity .

Hence,

$\sf{ V = \dfrac{ dx }{ dt } } \\ \\ \sf{ \implies { V = \dfrac{ d }{ dt } ( 18t + 5t^2 ) }} \\ \\ \sf{ \implies { V = \dfrac{ d }{ dt } ( 18 t ) + \dfrac{ d}{ dt } ( 5 t ^ 2 ) }} \\ \\ \sf{ \implies { V = 10t + 18 }}$

Hence , the required equation for the velocity of the particle becomes -

V = 10t + 18

t¹ = 2 second

Velocity = 38 unit / second

When

t² = 4 second

Velocity = 58 unit / second

Average acceleration -

=> ( V2 - v1 ) / ( t2 - t1 )

=> ( 58 - 38 ) / ( 4 - 2 )

=> 20 / 2

=>10 unit / second ² ........ [ Answer ]

Answer 2

GiVeN :-

The displacement of a particle moving along x-axis is given by x = 18t + 5².

To DeTeRmInE :-

The average acceleration during the interval t₁ = 2 secs and t₂ = 4 secs.

SoLuTiOn :-

For acceleration, we first need to find the velocity.

Differentiating displacement :-

$\sf{v=\dfrac{dx}{dt}=\dfrac{d}{dt} 18t+5t^2=\dfrac{d}{dt}(18t)+\dfrac{d}{dt}5t^2 }\\\\\displaystyle{\sf{\implies v= 18+2\times5^{2-1}}}\\\\\displaystyle{\sf{\implies v= 18+10t}}$

Hence, v = 18 + 10t.

Method - 1

Now, differentiating velocity :-

$\displaystyle{\sf{a=\frac{dv}{dt}=\frac{d}{dt}(18+10t)=\frac{d}{dt}(18)+\frac{d}{dt}(10t) }}\\\\\displaystyle{\sf{\implies a=0+10}}\\\\\boxed{\displaystyle{\sf{\implies a=10\ units/s^2}}}$

Therefore, the average acceleration is 10 m/s².

$\rule{160}2$

Method - 2 (more appropriate for this question)

→ v = 18 + 10t

At t = 2 secs,

v₁ = 18 + 10(2)

⇒v₁ = 18 + 20

⇒v₁ = 38 unit/s

At t = 4 secs,

v₂ = 18 + 10(4)

⇒v₂ = 18 + 40

⇒v₂ = 58 unit/s

$\displaystyle{\sf{a=\frac{v_2-v_1}{t_2-t_1} }}\\\\\displaystyle{\sf{\implies a=\frac{58-38}{4-2} }}\\\\\displaystyle{\sf{\implies a=\frac{20}{2} }}\\\\\boxed{\displaystyle{\sf{\implies a=10\ unit/s^2} }}}$

Therefore, the average acceleration is 10 m/s².