1/2at^2-vt+d=0

Or

t=(v+√v^2-2ad)/a

How is the value of t get from 1st equation

From hc verma

To derive the expression for time \( t \) from the equation \( \frac{1}{2}at^2 - vt + d = 0 \), we can recognize that this is a quadratic equation in the standard form \( Ax^2 + Bx + C = 0 \). Here, the coefficients correspond to \( A = \frac{1}{2}a \), \( B = -v \), and \( C = d \). We can apply the quadratic formula, which is given by \( t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \). Let's break it down step by step.

Identifying Coefficients

In our equation, we first identify the coefficients:

• A: \( \frac{1}{2}a \)
• B: \( -v \)
• C: \( d \)

Applying the Quadratic Formula

The quadratic formula is used to find the values of \( t \) that satisfy the equation. Plugging in our values, we get:

• Step 1: Replace \( A \), \( B \), and \( C \) in the formula:
• Step 2: Calculate \( B^2 - 4AC \):

Calculating \( B^2 \):

\( (-v)^2 = v^2 \)

Calculating \( 4AC \):

\( 4 \cdot \frac{1}{2}a \cdot d = 2ad \)

Now, we can find \( B^2 - 4AC \):

\( v^2 - 2ad \)

Final Steps in the Formula

Substituting back into the formula, we have:

\( t = \frac{-(-v) \pm \sqrt{v^2 - 2ad}}{2 \cdot \frac{1}{2}a} = \frac{v \pm \sqrt{v^2 - 2ad}}{a} \)

Since we're looking for the positive root (time cannot be negative in this context), we focus on the positive sign:

Thus, we arrive at:

\( t = \frac{v + \sqrt{v^2 - 2ad}}{a} \)

Understanding the Variables

This formula provides the time \( t \) in terms of the initial velocity \( v \), acceleration \( a \), and distance \( d \). It essentially tells you how long it will take for an object to travel a certain distance while considering its initial velocity and the acceleration acting on it.

By analyzing the components of the equation and applying the quadratic formula, we have derived the expression for time effectively. This approach can be applied in various physics problems that involve motion, enhancing your understanding of motion under constant acceleration. If you have further questions about the specifics or applications, feel free to ask!