Two particles move along x axis. the position of particle 1 is given by x = 6.00 $t^{2}$ + 3.00t + 2.00 (in meters and seconds); the acceleration of particle 2 is given by a = -8.00t ( in meters per second squared and seconds) and, at t = 0, its velocity is 19 m/s. When the velocities of the particles match, what is their velocity?

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Solution

The correct option is A
1

Particle 1 x = $6. t^{2} + 3 t + 2$

So, to find velocity we need to differentiate

As slope of x - t graph is velocity

⇒ $\frac{d x}{d t}$ = v = 12t + 3 .......(1)

Particle 2 a = - 8t

Acceleration if a = $\frac{d v}{d t} = - 8 t$

Rate of change = $\int d v = \int - 8 t d t$

It's given that at t = 0 velocity of particle is 19 m/s

So at time t let velocity be v

So limits are

$\int_{19}^{v}$ dv = $\int_{0}^{t} - 8 t d t$

$v]_{19}^{v} = \frac{- 8 t^{2}}{2}]_{0}^{t}$

v = 19 - 4 $t^{2}$

v = $- 4 t^{2} + 19$ .......... (2)

Let's assume at time $t_{0}$ , the velocity of both particles are some so we get

Equation (1) = equation (2)

$12 t_{0} + 3 = - 4 t_{0}^{2} + 19$

$4 t_{0}^{2} + 3 t_{0} - 4 = 0$

$(t_{0} + 4) (t_{0} - 1) = 0$

$t_{0} = - 4$

So $t_{0} = 1 s e c$

Since , after motion or t = 0 so t cannot be begative

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