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(I) A sports car accelerates from rest to 95 km/h in 4.3 s. What is its average acceleration in m/s\(^2\)?

Short Answer

Expert verified
The average acceleration is 6.14 m/s\(^2\).

Step by step solution

01

Convert Speed from km/h to m/s

The final speed is given in km/h, which must be converted to m/s. Use the conversion factor \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). The final speed in m/s is \(95 \times \frac{1}{3.6} = 26.39 \text{ m/s}\).
02

Identify Initial and Final Velocities

The initial velocity \(v_i\) is 0 m/s (since the car starts from rest). The final velocity \(v_f\) is the value calculated in the previous step: \(26.39 \text{ m/s}\).
03

Identify Time Duration

The time duration \(t\) over which the acceleration occurs is 4.3 seconds.
04

Apply the Average Acceleration Formula

The formula for average acceleration \(a\) is \(a = \frac{v_f - v_i}{t}\). Substitute the known values: \(v_f = 26.39 \text{ m/s}\), \(v_i = 0 \text{ m/s}\), and \(t = 4.3 \text{ s}\).
05

Calculate the Average Acceleration

Substitute the values into the formula: \(a = \frac{26.39 \, \text{m/s} - 0 \, \text{m/s}}{4.3 \, \text{s}} = 6.14 \, \text{m/s}^2\). The average acceleration is therefore 6.14 m/s\(^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is the branch of physics that explores the motion of objects without considering the forces causing that motion. It focuses on descriptions of movement in terms of displacement, velocity, and acceleration. When considering a problem involving kinematics, such as a sports car accelerating, one must identify these parameters:
  • Displacement: This refers to the distance moved in a particular direction.
  • Velocity: The rate of change of displacement. It reflects both speed and direction.
  • Acceleration: The rate of change of velocity. This can be positive (speeding up) or negative (slowing down).
In our exercise of the sports car accelerating from rest to a specified speed, kinematics helps us determine the vehicle's average acceleration. This involves careful attention to initial and final velocities, and the time over which the motion occurs.
Velocity Conversion
Velocity conversion is a crucial step when solving problems in kinematics involving different unit systems. Frequently, you may find speed measurements presented in non-SI units such as kilometers per hour (km/h).
In such cases, you need to convert these velocities into meters per second (m/s), the SI unit for velocity, to apply kinematic equations correctly. This conversion is essential because using consistent units simplifies the mathematical process and reduces errors.
  • To convert from km/h to m/s, divide the speed by 3.6. This factor comes from the conversion of kilometers to meters and hours to seconds.
  • Example: A speed of 95 km/h is converted to m/s by computing: \[95 \times \frac{1}{3.6} = 26.39 \text{ m/s}\]
By accurately converting velocities, you ensure that all calculations are aligned with the principles of physics and yield correct results.
Motion Equations
Motion equations form the backbone of kinematic calculations. These equations allow us to describe various aspects of motion mathematically. For the problem at hand, we derive the average acceleration using a fundamental motion equation:
The formula for average acceleration is expressed as:\[a = \frac{v_f - v_i}{t}\]
In this equation:
  • \(v_f\) is the final velocity.
  • \(v_i\) is the initial velocity.
  • \(t\) is the time taken for the change in velocity.
By substituting these values into the equation, we can calculate the vehicle's average acceleration.
From our example, we find:
  • Initial velocity (\(v_i\)): 0 m/s (the car starts from a complete stop)
  • Final velocity (\(v_f\)): 26.39 m/s (converted from km/h)
  • Time (\(t\)): 4.3 seconds
Substituting these values into the formula:\[a = \frac{26.39 \, \text{m/s} - 0 \, \text{m/s}}{4.3 \, \text{s}} = 6.14 \, \text{m/s}^2\]
This results in the average acceleration of the car being 6.14 m/s², illustrating how motion equations provide a structured path to solve kinematic problems.

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Most popular questions from this chapter

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