An insect moving along straight line, (without returning) travels in every second distance equal to time elapsed. Assuming acceleration to be constant and the insect starts at the=0, the magnitude of initial velocity will be

To determine the initial velocity of the insect moving along a straight line with constant acceleration, we can use the relationship between distance, time, and velocity. Given that the distance traveled in each second is equal to the time elapsed, we can derive the initial velocity from this information.

Understanding the Motion

Let's break down the problem. The insect starts from rest at position \( s = 0 \) and travels a distance that is equal to the time elapsed. This means that in the first second, it travels 1 meter, in the second second, it travels 2 meters, and so on. If we denote the time elapsed as \( t \), the distance traveled in \( t \) seconds can be expressed as:

• Distance in 1 second: 1 meter
• Distance in 2 seconds: 2 meters
• Distance in 3 seconds: 3 meters

In general, the distance \( s \) traveled after \( t \) seconds can be represented as:

s = t

Applying Kinematic Equations

For an object under constant acceleration, the distance traveled can also be described by the kinematic equation:

s = ut + (1/2)at²

Where:

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

Setting Up the Equation

Since we know that \( s = t \), we can substitute this into the kinematic equation:

t = ut + (1/2)at²

Rearranging the Equation

Now, let's rearrange this equation to isolate the initial velocity \( u \). First, we can rewrite it as:

t - ut - (1/2)at² = 0

Finding the Acceleration

To find the acceleration, we can look at the distance traveled in the second second. The distance traveled in the second second is given by:

Distance in second second = s(2) - s(1) = 2 - 1 = 1 meter

This indicates that the insect is accelerating. Since the distance traveled in the second second is equal to the distance traveled in the first second, we can conclude that the acceleration \( a \) must be 1 m/s².

Substituting Back

Now we can substitute \( a = 1 \) m/s² back into our equation:

t = ut + (1/2)(1)t²

Which simplifies to:

t = ut + (1/2)t²

Solving for Initial Velocity

Now, we can factor out \( t \) from the equation:

t = t(u + (1/2)t)

Since \( t \) cannot be zero, we can divide both sides by \( t \):

1 = u + (1/2)t

Rearranging gives us:

u = 1 - (1/2)t

Finding Initial Velocity at t = 0

At the very start, when \( t = 0 \):

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

Thus, the magnitude of the initial velocity \( u \) is:

Initial Velocity = 1 m/s

Summary

In summary, the insect starts from rest and accelerates at a constant rate. Given the conditions of the problem, we find that the initial velocity is 1 m/s. This example illustrates how we can apply kinematic equations to analyze motion under constant acceleration, leading us to understand the relationship between distance, time, and velocity effectively.