Problem 25 - Fundamentals of Physics

Question

an clectric vehicle starts from rest and accelerates at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) in a straight line until it reaches a speed of \(20 \mathrm{~m} / \mathrm{s} .\) The vchicle then slows at a constant rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?

Step-by-Step Solution

Answer

(a) 30 seconds; (b) 300 meters.

Step 1: Calculating Time to Reach Final Speed

To find the time it takes for the vehicle to reach a speed of 20 m/s from rest with an acceleration of 2.0 m/s², we use the formula \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity (0 m/s), \( a \) is the acceleration, and \( t \) is the time.\(ewline v = 20 \text{ m/s}, u = 0 \text{ m/s}, a = 2.0 \text{ m/s}^2 \)\(ewline 20 = 0 + 2.0t \)\(ewline t = \frac{20}{2.0} = 10 \text{ seconds} \).

Step 2: Calculating Time to Stop

Now, calculate the time to stop the vehicle from 20 m/s to 0 m/s with a deceleration of 1.0 m/s² using the formula \( v = u + at \).\(ewline v = 0 \text{ m/s}, u = 20 \text{ m/s}, a = -1.0 \text{ m/s}^2 \)\(ewline 0 = 20 - 1.0t \)\(ewline t = \frac{20}{1.0} = 20 \text{ seconds} \).

Step 3: Total Time Elapsed

Add the time calculated in Step 1 and Step 2 to get the total time from start to stop.\(ewline t_{\text{total}} = 10 + 20 = 30 \text{ seconds} \).

Step 4: Calculating Distance During Acceleration

Use the formula \( s = ut + \frac{1}{2}at^2 \) to calculate the distance traveled during the acceleration phase.\(ewline u = 0 \text{ m/s}, a = 2.0 \text{ m/s}^2, t = 10 \text{ seconds} \)\(ewline s = 0 \cdot 10 + \frac{1}{2} \cdot 2.0 \cdot (10)^2 \)\(ewline s = 100 \text{ meters} \).

Step 5: Calculating Distance During Deceleration

Use the same formula for the deceleration phase.\(ewline u = 20 \text{ m/s}, a = -1.0 \text{ m/s}^2, t = 20 \text{ seconds} \)\(ewline s = 20 \cdot 20 + \frac{1}{2} \cdot (-1.0) \cdot (20)^2 \)\(ewline s = 400 - 200 = 200 \text{ meters} \).

Step 6: Total Distance Traveled

Add the distances from the acceleration and deceleration phases to find the total distance.\(ewline s_{\text{total}} = 100 + 200 = 300 \text{ meters} \).

Key Concepts

Acceleration

Acceleration is the process of an object increasing its speed over time. In kinematics, acceleration is a vector quantity, meaning it has both a magnitude and a direction. It is typically measured in meters per second squared (m/s²). When a vehicle accelerates, it means its velocity is increasing each second.
For example, in our exercise, the electric vehicle starts from rest and accelerates at a rate of 2.0 m/s². This means that every second, the speed of the vehicle increases by 2 meters per second. Acceleration can be calculated using the formula:

• \(v = u + at\)
• \(v\) is the final velocity
• \(u\) is the initial velocity
• \(a\) is the acceleration
• \(t\) is the time

This formula shows how long it takes for an object to reach a certain speed given a constant acceleration. Understanding this concept is crucial in analyzing motion, whether it starts from rest or any initial speed.

Deceleration

Deceleration is essentially negative acceleration, meaning the object is slowing down. It occurs when a vehicle or any moving object decreases its velocity over time. In the context of our exercise, it refers to the phase where the electric vehicle reduces its speed from 20 m/s to a complete stop at a rate of -1.0 m/s².
This rate implies that the vehicle's speed decreases by 1 meter per second every second until it stops. The formula for deceleration is similar to that of acceleration but with the acceleration value being negative:

• \(v = u + at\)
• \(v\) is the final velocity, in this case, 0 m/s for a stop
• \(u\) is the initial velocity before deceleration begins
• \(a\) is negative acceleration (deceleration)
• \(t\) is the time

Understanding deceleration is key to predicting how objects slow down and eventually come to a stop, a fundamental part of kinematics and everyday life scenarios.

Velocity

Velocity is the speed of an object in a specific direction. Unlike simple speed, which is a scalar quantity, velocity is a vector, meaning it takes direction into account. It's measured in meters per second (m/s). To understand motion in kinematics, distinguishing between velocity and speed is essential.
In the exercise example, the electric vehicle changes its velocity from 0 m/s (rest) to 20 m/s during its acceleration phase. The velocity of the vehicle is manipulated directly by the forces acting on it, covering both increase in speed during acceleration and decrease during deceleration.
It's important to note that velocity can change even if the speed remains constant, provided there's a change in direction. However, in cases of motion in a straight line, velocity changes are typically changes in speed. In equations:

• \(v = u + at\) is used for velocity with uniform acceleration.

This forms the basis for understanding how speed and velocity interact under different forces, pivotal for solving kinematic problems.

Distance

Distance refers to the total path traveled by an object during motion, irrespective of direction. In our kinematic analysis, it's derived from both acceleration and deceleration phases. Distance is a scalar quantity and is typically measured in meters.
During the exercise, we calculate the distance covered by the electric vehicle in two phases: while accelerating and while decelerating.

• In the acceleration phase, the vehicle covers 100 meters.
• While decelerating, it covers an additional 200 meters.

The formula used to find the distance in both phases is:

• \(s = ut + \frac{1}{2}at^2\)
• \(s\) is the distance
• \(u\) is the initial velocity
• \(a\) is the acceleration (positive or negative for deceleration)
• \(t\) is the time

By understanding and using this formula, we calculate the total distance traveled as 300 meters in the example. Grasping the concept of distance is crucial for solving motion problems, letting us comprehend the actual path taken by moving objects.