Question

The position of a particle of a body moving in a straight line is x=2t^2+2t+9 where did in metre and tis time in second .the velocity v(v=DX/dt) of the body at t=1sec​

Answer 1

Given :

• Distance of a body is given by x = 2t² + 2t + 9

To Find :

• Find velocity of particle at t = 1 s

Solution :

We're given that the position of a body is given by the equation 2t² + 2t + 9. Where,

• t is time interval

And we've to find velocity of the particle when time is 1 second. We will use here differentiation.

$\implies \sf{x \: = \: 2t^2 \: + \: 2t \: + \: 9} \\ \\ \implies \sf{\dfrac{dx}{dt} \: = \: \dfrac{d(2t^2 \: + \: 2t \: + \: 9)}{dt}} \\ \\ \implies \sf{v \: = \: 2 \times \: 2(t) \: + \: 2 \: + \: 0} \\ \\ \implies \sf{v \: = \: 4t \: + \: 2} \\ \\ \longrightarrow \underline{\boxed{\sf{v \: = \: 4t \: + \: 2 \: ms^{-1}}}}$

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Now, put time (t) as 1 second

$\implies \sf{v \: = \: 4(1) \: + \: 2} \\ \\ \implies \sf{v \: = \: 4 \: + \: 2} \\ \\ \implies \sf{v \: = \: 6}$

$\therefore$ Velocity of the particle at t = 1 sec is 6 m/s

Answer 2

the correct answer is 6m/s