Question

Two particles a and b are moving in xy-plane. their positions vary with time t according to relation: xa(t) = 3t, xâb(t) = 6 ya(t) = t, ybâ(t) = 2 + 3t2 distance between two particles at t = 1 is: (1) 5 (2) 3 â(3) 4

Answer 1

Answer:

The distance between two particles is 5 m.

1 option is correct.

Explanation:

Given that,

Relation between position and time

$x_a {t}=3t$

$x_b {t}=6$

$y_b {t}=t$

$y_b {t}=2+3t^2$

The position of the particle at t = 1,

$x_{a}=3$

$x_{b}=6$

$y_{b}=1$

$y_{b}=5$

The coordinates of the particles

$x_{a},x_{b}=3,6$

$y_{a},y_{b}=1,5$

The distance between two particles is

$d=\sqrt{(x_{b}-x_{a})^2+(y_{b}-y_{a})^2}$

$d=\sqrt{(6-3)^2+(5-1)^2}$

$d=5 m$

Hence, The distance between two particles is 5 m.

Answer 2

Answer:

d = 12 if t is 2

Explanation:

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