A Car Moving South Speeds Up From $10 , \text{m/s}$ To $40 , \text{m/s}$ In 15 Seconds. What Is The Car's Acceleration?A. 2 M/s 2 2 \, \text{m/s}^2 2 M/s 2 B. 15 M/s 2 15 \, \text{m/s}^2 15 M/s 2 C. 30 M/s 2 30 \, \text{m/s}^2 30 M/s 2 D.

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Understanding the Problem

To find the car's acceleration, we need to use the concept of acceleration, which is defined as the rate of change of velocity. In this case, the car's velocity changes from $10 , \text{m/s}$ to $40 , \text{m/s}$ in 15 seconds. We can use the formula for acceleration, which is:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time over which the change occurs.

Calculating the Change in Velocity

The change in velocity ($\Delta v$) is the difference between the final velocity and the initial velocity:

Ξ”v=vfβˆ’vi\Delta v = v_f - v_i

where $v_f$ is the final velocity ($40 , \text{m/s}$) and $v_i$ is the initial velocity ($10 , \text{m/s}$).

Ξ”v=40 m/sβˆ’10 m/s=30 m/s\Delta v = 40 \, \text{m/s} - 10 \, \text{m/s} = 30 \, \text{m/s}

Calculating the Acceleration

Now that we have the change in velocity, we can plug it into the formula for acceleration:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $\Delta t$ is the time over which the change occurs (15 seconds).

a=30 m/s15 s=2 m/s2a = \frac{30 \, \text{m/s}}{15 \, \text{s}} = 2 \, \text{m/s}^2

Conclusion

The car's acceleration is $2 , \text{m/s}^2$. This means that the car is accelerating at a rate of $2 , \text{m/s}^2$, which is a relatively slow acceleration.

Comparison with Other Options

Let's compare our answer with the other options:

  • A. $2 , \text{m/s}^2$: This is our calculated answer.
  • B. $15 , \text{m/s}^2$: This is not our calculated answer.
  • C. $30 , \text{m/s}^2$: This is not our calculated answer.
  • D. (no answer provided): This is not an option.

Final Answer

The final answer is A. $2 , \text{m/s}^2$.

Discussion

This problem requires the application of the concept of acceleration, which is a fundamental concept in physics. The formula for acceleration is:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time over which the change occurs.

In this problem, we were given the initial and final velocities, as well as the time over which the change occurs. We used the formula for acceleration to calculate the car's acceleration.

Tips and Tricks

  • Make sure to use the correct formula for acceleration.
  • Use the correct units for velocity and time.
  • Check your calculations carefully to avoid errors.

Related Problems

  • A car moving north speeds up from $20 , \text{m/s}$ to $60 , \text{m/s}$ in 10 seconds. What is the car's acceleration?
  • A car moving east speeds up from $30 , \text{m/s}$ to $90 , \text{m/s}$ in 15 seconds. What is the car's acceleration?

Conclusion

In conclusion, the car's acceleration is $2 , \text{m/s}^2$. This problem requires the application of the concept of acceleration, which is a fundamental concept in physics. The formula for acceleration is:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time over which the change occurs.

Understanding the Problem

To find the car's acceleration, we need to use the concept of acceleration, which is defined as the rate of change of velocity. In this case, the car's velocity changes from $10 , \text{m/s}$ to $40 , \text{m/s}$ in 15 seconds. We can use the formula for acceleration, which is:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time over which the change occurs.

Calculating the Change in Velocity

The change in velocity ($\Delta v$) is the difference between the final velocity and the initial velocity:

Ξ”v=vfβˆ’vi\Delta v = v_f - v_i

where $v_f$ is the final velocity ($40 , \text{m/s}$) and $v_i$ is the initial velocity ($10 , \text{m/s}$).

Ξ”v=40 m/sβˆ’10 m/s=30 m/s\Delta v = 40 \, \text{m/s} - 10 \, \text{m/s} = 30 \, \text{m/s}

Calculating the Acceleration

Now that we have the change in velocity, we can plug it into the formula for acceleration:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $\Delta t$ is the time over which the change occurs (15 seconds).

a=30 m/s15 s=2 m/s2a = \frac{30 \, \text{m/s}}{15 \, \text{s}} = 2 \, \text{m/s}^2

Conclusion

The car's acceleration is $2 , \text{m/s}^2$. This means that the car is accelerating at a rate of $2 , \text{m/s}^2$, which is a relatively slow acceleration.

Comparison with Other Options

Let's compare our answer with the other options:

  • A. $2 , \text{m/s}^2$: This is our calculated answer.
  • B. $15 , \text{m/s}^2$: This is not our calculated answer.
  • C. $30 , \text{m/s}^2$: This is not our calculated answer.
  • D. (no answer provided): This is not an option.

Final Answer

The final answer is A. $2 , \text{m/s}^2$.

Discussion

This problem requires the application of the concept of acceleration, which is a fundamental concept in physics. The formula for acceleration is:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time over which the change occurs.

In this problem, we were given the initial and final velocities, as well as the time over which the change occurs. We used the formula for acceleration to calculate the car's acceleration.

Tips and Tricks

  • Make sure to use the correct formula for acceleration.
  • Use the correct units for velocity and time.
  • Check your calculations carefully to avoid errors.

Related Problems

  • A car moving north speeds up from $20 , \text{m/s}$ to $60 , \text{m/s}$ in 10 seconds. What is the car's acceleration?
  • A car moving east speeds up from $30 , \text{m/s}$ to $90 , \text{m/s}$ in 15 seconds. What is the car's acceleration?

Q&A

Q: What is the formula for acceleration?

A: The formula for acceleration is:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time over which the change occurs.

Q: What is the change in velocity?

A: The change in velocity ($\Delta v$) is the difference between the final velocity and the initial velocity:

Ξ”v=vfβˆ’vi\Delta v = v_f - v_i

where $v_f$ is the final velocity and $v_i$ is the initial velocity.

Q: What is the time over which the change occurs?

A: The time over which the change occurs ($\Delta t$) is the time it takes for the velocity to change from the initial velocity to the final velocity.

Q: How do I calculate the acceleration?

A: To calculate the acceleration, you need to plug the change in velocity and the time over which the change occurs into the formula for acceleration:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

Q: What is the unit of acceleration?

A: The unit of acceleration is $\text{m/s}^2$.

Q: What is the final answer to the problem?

A: The final answer to the problem is A. $2 , \text{m/s}^2$.

Q: What is the car's acceleration?

A: The car's acceleration is $2 , \text{m/s}^2$.

Q: What is the difference between velocity and acceleration?

A: Velocity is the rate of change of position, while acceleration is the rate of change of velocity.

Q: What is the relationship between velocity and acceleration?

A: The relationship between velocity and acceleration is that acceleration is the rate of change of velocity.

Q: How do I use the formula for acceleration to solve problems?

A: To use the formula for acceleration to solve problems, you need to plug in the values for the change in velocity and the time over which the change occurs.

Conclusion

In conclusion, the car's acceleration is $2 , \text{m/s}^2$. This problem requires the application of the concept of acceleration, which is a fundamental concept in physics. The formula for acceleration is:

a=Ξ”vΞ”ta = \frac{\Delta v}{\Delta t}

where $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time over which the change occurs.