A Wodden plank of mass m and dimensions L*B floats over a very shallow river of thicknessT and vicosity η . at t=0 the speed of plank is v.a car is at rest on the bank at this instant the plank is adjacent to the car which starts to move with a constant acceleration.if after sometime when the speed of the plank is v/2 the car catches up with the plank(the car moves with a constant acceleration)find the speed of the car at this instant when the car is crossing the plank.

(A)3v/2

(B)vηLB/2T

(C)v/ln2

(D)3v/4

To solve this problem, we need to analyze the motion of both the wooden plank and the car. The plank is floating on a river, and at the initial moment, it has a speed of "v." The car starts from rest and accelerates to catch up with the plank when the plank's speed reduces to "v/2." Let's break this down step by step.

Understanding the Motion of the Plank

The plank is floating on water, and its motion is influenced by the viscosity of the water and the thickness of the river. However, since the problem states that the plank is initially moving at speed "v" and later at "v/2," we can assume that the plank's motion is primarily horizontal and that it experiences a deceleration due to the viscous drag from the water.

Deceleration of the Plank

Let’s denote the deceleration of the plank as "a_p." The relationship between the forces acting on the plank and its deceleration can be expressed through Newton's second law. The viscous force acting on the plank can be modeled as:

• F_viscous = η * A * (dv/dt), where A = L * B (the area of the plank).

Since the plank is decelerating, we can express this as:

• m * a_p = η * A * (v - v/2) / t,

From this, we can derive the deceleration of the plank over time until it reaches the speed of "v/2."

Analyzing the Car's Motion

The car starts from rest and accelerates with a constant acceleration "a_c." The distance covered by the car when it catches up with the plank can be expressed using the kinematic equation:

• d = 0.5 * a_c * t^2.

At the same time, the plank is moving, and we can express its distance traveled as:

• d_plank = v * t - 0.5 * a_p * t^2.

Setting the Distances Equal

At the moment the car catches up with the plank, the distances traveled by both must be equal:

• 0.5 * a_c * t^2 = v * t - 0.5 * a_p * t^2.

Rearranging this gives us:

• (0.5 * a_c + 0.5 * a_p) * t^2 = v * t.

Finding the Speed of the Car

Now, we need to find the speed of the car at the moment it crosses the plank. The speed of the car at that instant can be calculated using the formula:

• v_c = a_c * t.

From our earlier equation, we can express "a_c" in terms of "v" and "t." After some algebraic manipulation, we can find that:

• a_c = (2v / t) - a_p.

Substituting this back into the equation for the speed of the car gives us:

• v_c = (2v / t) * t - a_p * t = 2v - a_p * t.

Final Calculation

To find the exact value of "v_c" when the car crosses the plank, we need to substitute the value of "a_p" based on the viscosity and dimensions of the plank. After substituting and simplifying, we can arrive at the final expression for the speed of the car.

After performing these calculations, we find that the speed of the car when it crosses the plank is:

• (A) 3v/2.

Thus, the correct answer is (A) 3v/2. This solution illustrates the interplay between the forces acting on both the plank and the car, as well as the kinematic relationships governing their motions.