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Grade 12th passMechanics

A particle moves with a velocity of 2m/s in a straight line with a negative acceleration of 0.1m/s2 . Time taken to traverse a distance of 1.5m is

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8 Years agoGrade 12th pass
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ApprovedApproved Tutor Answer1 Year ago

To find the time taken for the particle to traverse a distance of 1.5 meters while moving with an initial velocity of 2 m/s and experiencing a negative acceleration of 0.1 m/s², we can use one of the equations of motion. Specifically, we will apply the second equation of motion, which relates distance, initial velocity, time, and acceleration.

Understanding the Variables

In this scenario, we have the following variables:

  • Initial velocity (u): 2 m/s
  • Acceleration (a): -0.1 m/s² (negative indicates deceleration)
  • Distance (s): 1.5 m
  • Time (t): unknown

Applying the Equation of Motion

The equation we will use is:

s = ut + (1/2)at²

Substituting the known values into the equation:

1.5 = (2)t + (1/2)(-0.1)t²

Rearranging the Equation

Let's simplify this equation step by step:

1.5 = 2t - 0.05t²

Rearranging gives us:

0.05t² - 2t + 1.5 = 0

Solving the Quadratic Equation

This is a standard quadratic equation in the form of at² + bt + c = 0, where:

  • a: 0.05
  • b: -2
  • c: 1.5

We can solve this using the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

Substituting the values:

t = (2 ± √((-2)² - 4(0.05)(1.5))) / (2 * 0.05)

Calculating the discriminant:

t = (2 ± √(4 - 0.3)) / 0.1

t = (2 ± √3.7) / 0.1

Now, calculating √3.7 gives approximately 1.923. Thus:

t = (2 ± 1.923) / 0.1

Finding the Two Possible Values for Time

This results in two potential solutions:

  • t₁ = (2 + 1.923) / 0.1 = 39.23 seconds
  • t₂ = (2 - 1.923) / 0.1 = 0.77 seconds

Interpreting the Results

Since the particle is decelerating, the time taken to cover 1.5 meters will be the smaller value, which is:

t = 0.77 seconds

In summary, the time taken for the particle to traverse a distance of 1.5 meters, given its initial velocity and negative acceleration, is approximately 0.77 seconds. This example illustrates how to apply the equations of motion effectively to solve real-world problems involving linear motion.