2 objects are moving along same line.they cross a point A with acc.of a and 2a and velocity 2u and u at t=0.The distance travelled by the object when one overtaked the other is?

To find the distance traveled by the two objects when one overtakes the other, we can use the equations of motion. Let's break down the problem step by step.

Understanding the Motion of Each Object

We have two objects moving along the same line:

• Object 1: Initial velocity = 2u, acceleration = a
• Object 2: Initial velocity = u, acceleration = 2a

Equations of Motion

For uniformly accelerated motion, the distance traveled by an object can be calculated using the formula:

s = ut + (1/2)at²

Where:

• s = distance traveled
• u = initial velocity
• a = acceleration
• t = time

Calculating Distance for Each Object

Let's denote the time taken until one object overtakes the other as t. We can calculate the distance traveled by each object at that time.

Distance Traveled by Object 1

Using the formula for Object 1:

s₁ = (2u)t + (1/2)(a)t²

Distance Traveled by Object 2

For Object 2, we apply the same formula:

s₂ = ut + (1/2)(2a)t²

This simplifies to:

s₂ = ut + at²

Setting the Distances Equal

Since one object overtakes the other when they have traveled the same distance from point A, we set s₁ equal to s₂:

(2u)t + (1/2)(a)t² = ut + at²

Simplifying the Equation

Rearranging the equation gives:

(2u)t - ut + (1/2)(a)t² - at² = 0

This simplifies to:

ut - (1/2)(a)t² = 0

Factoring Out t

Factoring out t from the equation, we have:

t( u - (1/2)at ) = 0

This gives us two solutions: t = 0 (the initial time) or:

u = (1/2)at

Finding Time of Overtaking

From the equation u = (1/2)at, we can solve for t:

t = (2u)/a

Calculating the Distance Traveled

Now, we can substitute this value of t back into the distance formula for either object. Let’s use Object 1:

s₁ = (2u)((2u)/a) + (1/2)(a)((2u)/a)²

This simplifies to:

s₁ = (4u²/a) + (1/2)(a)(4u²/a²)

s₁ = (4u²/a) + (2u²/a)

s₁ = (6u²/a)

Final Result

The distance traveled by either object when one overtakes the other is:

s = (6u²/a)

This result shows how the initial velocities and accelerations of the objects influence the distance they travel before one overtakes the other. If you have any further questions or need clarification on any part of this process, feel free to ask!