Question

A long horizontal belt is moving from left to right

with a uniform speed of 2m/s. There are two ink

marks A and B on the belt 60 m apart. An insect

runs on the belt to and fro between A and B such

that its speed relative to the belt is constant and

equals 4m/s. When the insect is moving on the belt

in the direction of motion of the belt, its speed as

observed by a person standing on ground will be

(1) 6 m/s (2) 2 m/s (3) 1.5 m/s (4) 4 m/s​

Answer 1

Answer:

The uniform speed of the horizontal belt = 2 m/s.

The distance between A and B = 60 m.

The relative speed of the insect to that of the belt = 4 m/s.

We know that, Relative velocity_(I/B) = v_I - v_B

⇒ v_I - v_B = 4 m/s

⇒ v_I = 4 - v_B

⇒ v_I = 4 - 2

⇒ v_I = 2 m/s

Hence, Option (2) is corrrect.

Hence, The speed of the insect as  observed by a person standing on ground will be 2 m/s.

(Here, I is Insect and B is the belt.)

MORE INFORMATION -

Let x_A(0) and x_B(0) be the positions of the objects at time t = 0.

And x_A(t) and x_B(t) be the positions of the objects at time = t.

Now,

x_A(t) = x_A(0) + v_A•t

x_B(t) = x_B(0) + v_B•t

Displacement from object A to B = s_BA

⇒ s_BA = x_A(0) + v_A•t - x_B(0) + v_B•t

⇒ s_BA = [x_A(0) - x_B(0)]+ t(v_B - v_A)

We know, [x_A(0) - x_B(0)] = 0

So, s_BA = t(v_B - v_A)

Differentiating,

⇒ v_BA = d(s_BA)/dt

⇒ v_BA = v_B - v_A