Question

the acceleration - time graph of a particle moving along a straight line is given at what time velocity of particle becomes equal to its initial velocity

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Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{8 \:sec }}}$

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$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

An Acceleration time graph,

So, Area under the velocity time graph is zero.

Let the time t° the particle accuire the initial velocity,

Therefore the change in velocity of the particle in the time t is:

So, Area under the velocity time graph is zero.

Thus, Area of upper triangle (say AOB) = Area of lower triangle (say BCD)

1/2 OA×OB = 1/2 OC×CD

⇒OB² tan @ = CD² tan @

⇒OB² = CD²

⇒4 = t° - 4

⇒t° = 8 sec

Answer 2

Answer:

Option => 2

The time - 8 sec the particle acquires it's initial velocity.

Explanation:

Given :

The area under velocity graph represents the change in velocity.

To Find :

The time.

Solution :

★ Consider as -

At time t₀ the particle aquire the initial velocity.

Now,

It implies that the change in velocity of the particle in the time t₀ is zero,

Because, Area under av / s t graph is zero.

This whole statemet implies that,

The area of Δ AOB = Area of Δ BCD.

So,

$\sf \\ \\\implies \dfrac{1}{2} \times OA \times OB = \dfrac{1}{2} \times BC \times CD.\\ \\ \\ \implies OB\;tan\; \theta \times OB = CD\;tan\;\theta \times CD.\\ \\ \\\implies (OB)^2 = (CD)^2.\\ \\ \\ \implies OB = CD.\\ \\ \\\implies 4 = t_0 -4.\\ \\ \\\implies t_0 = 8$

Hence,

The time - 8 sec the particle acquires it's initial velocity.