A Body Starts From Rest And Moves With A Uniform Acceleration Of $6 , \text{m/s}^2$. What Distance Does It Cover In The Third Second?A. 15 M B. 18 M C. 27 M D. 30 M

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Introduction

When a body starts from rest and moves with a uniform acceleration, it is a classic example of uniformly accelerated motion. In this scenario, the body's velocity increases at a constant rate, and its position changes accordingly. Understanding the distance covered by the body in a given time frame is crucial in various fields, including physics, engineering, and sports. In this article, we will delve into the concept of uniformly accelerated motion and calculate the distance covered by a body in the third second, given a uniform acceleration of $6 , \text{m/s}^2$.

Uniformly Accelerated Motion

Uniformly accelerated motion is a fundamental concept in physics, where an object's acceleration remains constant over a period of time. The equation of motion for an object under uniform acceleration is given by:

s=ut+12at2s = ut + \frac{1}{2}at^2

where:

  • s$ is the distance traveled by the object

  • u$ is the initial velocity of the object

  • t$ is the time elapsed

  • a$ is the uniform acceleration

Calculating Distance Covered in the Third Second

To calculate the distance covered by the body in the third second, we need to use the equation of motion. Since the body starts from rest, the initial velocity $u$ is 0 m/s. The uniform acceleration $a$ is given as $6 , \text{m/s}^2$. We need to find the distance covered in the third second, which means we need to find the distance traveled from $t = 2$ seconds to $t = 3$ seconds.

Step 1: Find the Velocity at the End of the Second

First, we need to find the velocity of the body at the end of the second. We can use the equation:

v=u+atv = u + at

where:

  • v$ is the final velocity

  • u$ is the initial velocity (0 m/s)

  • a$ is the uniform acceleration ($6 \, \text{m/s}^2$)

  • t$ is the time elapsed (2 seconds)

Substituting the values, we get:

v=0+6×2v = 0 + 6 \times 2

v=12 m/sv = 12 \, \text{m/s}

Step 2: Find the Distance Traveled in the First Two Seconds

Next, we need to find the distance traveled by the body in the first two seconds. We can use the equation:

s=ut+12at2s = ut + \frac{1}{2}at^2

where:

  • s$ is the distance traveled

  • u$ is the initial velocity (0 m/s)

  • t$ is the time elapsed (2 seconds)

  • a$ is the uniform acceleration ($6 \, \text{m/s}^2$)

Substituting the values, we get:

s=0×2+12×6×22s = 0 \times 2 + \frac{1}{2} \times 6 \times 2^2

s=0+12×6×4s = 0 + \frac{1}{2} \times 6 \times 4

s=0+12s = 0 + 12

s=12 ms = 12 \, \text{m}

Step 3: Find the Distance Traveled in the Third Second

Now, we need to find the distance traveled by the body in the third second. We can use the equation:

s=ut+12at2s = ut + \frac{1}{2}at^2

where:

  • s$ is the distance traveled

  • u$ is the initial velocity (0 m/s)

  • t$ is the time elapsed (1 second)

  • a$ is the uniform acceleration ($6 \, \text{m/s}^2$)

However, we need to find the velocity at the beginning of the third second. We can use the equation:

v=u+atv = u + at

where:

  • v$ is the final velocity

  • u$ is the initial velocity (12 m/s)

  • a$ is the uniform acceleration ($6 \, \text{m/s}^2$)

  • t$ is the time elapsed (1 second)

Substituting the values, we get:

v=12+6×1v = 12 + 6 \times 1

v=18 m/sv = 18 \, \text{m/s}

Now, we can find the distance traveled in the third second:

s=ut+12at2s = ut + \frac{1}{2}at^2

where:

  • s$ is the distance traveled

  • u$ is the initial velocity (12 m/s)

  • t$ is the time elapsed (1 second)

  • a$ is the uniform acceleration ($6 \, \text{m/s}^2$)

Substituting the values, we get:

s=12×1+12×6×12s = 12 \times 1 + \frac{1}{2} \times 6 \times 1^2

s=12+12×6s = 12 + \frac{1}{2} \times 6

s=12+3s = 12 + 3

s=15 ms = 15 \, \text{m}

However, we need to add the distance traveled in the first two seconds to the distance traveled in the third second. We already found the distance traveled in the first two seconds to be 12 m. Therefore, the total distance traveled in the third second is:

s=12+15s = 12 + 15

s=27 ms = 27 \, \text{m}

Conclusion

In conclusion, a body starts from rest and moves with a uniform acceleration of $6 , \text{m/s}^2$. The distance covered by the body in the third second is 27 m.

Frequently Asked Questions

  • What is uniformly accelerated motion? Uniformly accelerated motion is a fundamental concept in physics, where an object's acceleration remains constant over a period of time.
  • How do you calculate the distance traveled by a body in uniformly accelerated motion? You can use the equation of motion: $s = ut + \frac{1}{2}at^2$
  • What is the distance covered by a body in the third second, given a uniform acceleration of $6 , \text{m/s}^2$? The distance covered by the body in the third second is 27 m.

References

  • Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of physics. John Wiley & Sons.
  • Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage Learning.

Introduction

In our previous article, we explored the concept of uniformly accelerated motion and calculated the distance covered by a body in the third second, given a uniform acceleration of $6 , \text{m/s}^2$. In this article, we will address some of the frequently asked questions related to uniformly accelerated motion and provide detailed explanations and examples.

Q&A

Q1: What is uniformly accelerated motion?

A1: Uniformly accelerated motion is a fundamental concept in physics, where an object's acceleration remains constant over a period of time.

Q2: How do you calculate the distance traveled by a body in uniformly accelerated motion?

A2: You can use the equation of motion: $s = ut + \frac{1}{2}at^2$

Q3: What is the significance of the equation of motion in uniformly accelerated motion?

A3: The equation of motion is a fundamental tool in uniformly accelerated motion, as it allows us to calculate the distance traveled by a body, given its initial velocity, acceleration, and time.

Q4: How do you calculate the velocity of a body in uniformly accelerated motion?

A4: You can use the equation: $v = u + at$

Q5: What is the relationship between velocity and acceleration in uniformly accelerated motion?

A5: In uniformly accelerated motion, the velocity of a body increases at a constant rate, which is equal to the acceleration.

Q6: How do you calculate the distance traveled by a body in the first second of uniformly accelerated motion?

A6: You can use the equation: $s = ut + \frac{1}{2}at^2$

Q7: What is the distance covered by a body in the third second, given a uniform acceleration of $6 , \text{m/s}^2$?

A7: The distance covered by the body in the third second is 27 m.

Q8: How do you calculate the distance traveled by a body in uniformly accelerated motion, given its initial velocity, acceleration, and time?

A8: You can use the equation of motion: $s = ut + \frac{1}{2}at^2$

Q9: What is the significance of the time elapsed in uniformly accelerated motion?

A9: The time elapsed is a crucial factor in uniformly accelerated motion, as it determines the distance traveled by a body.

Q10: How do you calculate the velocity of a body at the end of a given time in uniformly accelerated motion?

A10: You can use the equation: $v = u + at$

Examples

Example 1: A body starts from rest and moves with a uniform acceleration of $4 , \text{m/s}^2$. Calculate the distance covered by the body in the third second.

Solution: We can use the equation of motion: $s = ut + \frac{1}{2}at^2$

where:

  • s$ is the distance traveled

  • u$ is the initial velocity (0 m/s)

  • t$ is the time elapsed (3 seconds)

  • a$ is the uniform acceleration ($4 \, \text{m/s}^2$)

Substituting the values, we get:

s=0×3+12×4×32s = 0 \times 3 + \frac{1}{2} \times 4 \times 3^2

s=0+12×4×9s = 0 + \frac{1}{2} \times 4 \times 9

s=0+18s = 0 + 18

s=18 ms = 18 \, \text{m}

Example 2: A body starts from rest and moves with a uniform acceleration of $8 , \text{m/s}^2$. Calculate the distance covered by the body in the third second.

Solution: We can use the equation of motion: $s = ut + \frac{1}{2}at^2$

where:

  • s$ is the distance traveled

  • u$ is the initial velocity (0 m/s)

  • t$ is the time elapsed (3 seconds)

  • a$ is the uniform acceleration ($8 \, \text{m/s}^2$)

Substituting the values, we get:

s=0×3+12×8×32s = 0 \times 3 + \frac{1}{2} \times 8 \times 3^2

s=0+12×8×9s = 0 + \frac{1}{2} \times 8 \times 9

s=0+36s = 0 + 36

s=36 ms = 36 \, \text{m}

Conclusion

In conclusion, uniformly accelerated motion is a fundamental concept in physics, where an object's acceleration remains constant over a period of time. The equation of motion is a crucial tool in uniformly accelerated motion, as it allows us to calculate the distance traveled by a body, given its initial velocity, acceleration, and time. We have addressed some of the frequently asked questions related to uniformly accelerated motion and provided detailed explanations and examples.

Frequently Asked Questions

  • What is uniformly accelerated motion? Uniformly accelerated motion is a fundamental concept in physics, where an object's acceleration remains constant over a period of time.
  • How do you calculate the distance traveled by a body in uniformly accelerated motion? You can use the equation of motion: $s = ut + \frac{1}{2}at^2$
  • What is the significance of the equation of motion in uniformly accelerated motion? The equation of motion is a fundamental tool in uniformly accelerated motion, as it allows us to calculate the distance traveled by a body, given its initial velocity, acceleration, and time.

References

  • Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of physics. John Wiley & Sons.
  • Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage Learning.