Question

Two cars A and B are moving with same constant speed 5 m/s in opposite direction. The relative velocity of car A w.r.t car B is Zero 10 m/s 12 m/s 5 m/s

Answer 1

Explanation:

Suppose after time t the car A reaches the the car B then distance traveled by B i.e. x B

and the distance by A i.e. x A

should be such that x A

=x B+gap because A was behind B

or x A

=x B+80m ...(1)

Now as A moves with constant velocity so x

A

=10×t

and B is moving with acceleration and initially at rest (u=0) so x

B =0+a 0 t 2 =a 0 t 2

putting above value0s in equation-1 we get 10t=a0t 2 +80

If car A catches car B then we should get some value of t i.e. the root of above eqaution should be REAL.

So the discriminant should be positive i.e (10)2−4×a 0 ×80≥0 or a 0 ≤ 16/5 m/s 2

Answer 2

Answer: 10 m/s,

Given:

$\sf Velocity\:of\:car\:A=5\:m/s$

$\sf Velocity\:of\:car\:B=5\:m/s$

To find:

$\sf Relative\:velocity\:of\:car\:A\:w.r.t\:car\:B.$

Explanation:

$\sf According\:to\:the\:question,\\\:both\:cars\:are\:running\:in\:the\:opposite\:direction\:w.r.t\:each\:other.$

$\sf \therefore Relative\:velocity\:of\:car\A(V_A)=+5m/s$

$\sf Relative\:velocity\:of\:car\:B(V_B)=-5m/s$

$\sf Hence,$

$\boxed{\sf Relative\:velocity\:of\:A\:w.r.t\:B(V_{AB})=V_A-V_B}$

$\Rightarrow \sf \sf V_{AB}=5-(-5)$

$\Rightarrow \boxed{\sf V_{AB}=10\:m/s}$

Hence, the relative velocity of car A w.r.t car B is 10 m/s.

More to know:

Relative velocity is defined as the velocity of an object B in the rest frame of another object A.