Question

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?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A

1

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

So, to find velocity we need to differentiate

As slope of x - t graph is velocity

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

Particle 2 a = - 8t

Acceleration if a = $\frac{dv}{dt}=-8t$

Rate of change = $\int dv=\int -8tdt$

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 =${\int }_0^t-8tdt$

$v{]}_{19}^v=\frac{-8t^2}{2}{]}_0^t$

v = 19 - 4$t^2$

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

Let's assume at time $t_0$ , the velocity of both particles are some so we get

Equation (1) = equation (2)

$12t_0+3=-4t_0^2+19$

$4t_0^2+3t_0-4=0$

$(t_0+4)(t_0-1)=0$

$t_0=-4$

So $t_0=1sec$

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