Question

The position-time graph for two particles A and B
are straight lines inclined at angles of 45° and 60°
with time axis as shown in figure. The magnitude
of relative velocity of A w.r.t. B is
x(m)
B
А
60°
45°
t(s)​

Answer 1

The relative velocity of A w.r.t. B is (1 - √3) m/s

Explanation:

Given:

The inclination of position-time graph of particle A is 45°

The inclination of position-time graph of particle B is 60°

To find out:

The relative velocity of A w.r.t. B

Solution:

We know that the slope of the displacement time graph gives us the velocity

$v=\frac{dx}{dt}=\tan\theta$    Where $\theta$ is the inclination of the tangent to the displacement time curve at that instant)

Therefore,

Velocity of particle A w.r.t ground

$V_{A/G}=\tan 45^\circ$

$\implies V_{A/G}=1$ m/s

Velocity of particle B w.r.t ground

$V_{B/G}=\tan 60^\circ$

$\implies V_{B/G}=\sqrt{3}$ m/s

Therefore,

The relative velocity of A w.r.t. B

$V_{A/B}=V_{A/G}-V_{B/G}$

$\implies V_{A/B}=1-\sqrt{3}$ m/s

Hope this answer is helpful.

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