Kinematics Equations: Videos & Practice Problems

Understanding motion problems involves mastering the four key equations of motion, applicable only under constant acceleration. These equations relate variables such as initial velocity (v₀), final velocity (v), acceleration (a), displacement (Δx), and time (t). Recognizing the signs of acceleration is crucial; positive acceleration indicates increasing velocity when aligned with initial velocity. Conversely, negative acceleration can still result in increasing speed if the initial velocity is also negative. Organizing variables and selecting the appropriate equation is essential for solving complex motion scenarios effectively.

Kinematics Equations

Understanding motion problems involving acceleration is crucial in physics, particularly when applying the four key equations of motion, also known as kinematics equations. These equations are essential for analyzing uniformly accelerated motion, which occurs when acceleration remains constant. It's important to note that these equations can only be used under this condition.

The first equation is given by:

$$v = v_0 + at$$

where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. This equation can be derived from the average acceleration formula, which states that the average acceleration is equal to the change in velocity over the change in time.

The second equation is:

$$v^2 = v_0^2 + 2a\Delta x$$

This equation relates the squares of the velocities to acceleration and displacement, where \(\Delta x\) represents the change in position.

The third equation can be expressed in two forms:

$$\Delta x = v_0t + \frac{1}{2}at^2$$

or

$$\Delta x = \frac{(v_0 + v)}{2}t$$

In the first form, \(\Delta x\) is the displacement, while in the second form, it represents the average velocity multiplied by time. The average velocity is calculated as the mean of the initial and final velocities.

Each of these equations incorporates a combination of five key variables: final velocity (\(v\)), initial velocity (\(v_0\)), acceleration (\(a\)), time (\(t\)), and displacement (\(\Delta x\)). When solving motion problems, it is essential to identify which three of these five variables are known to select the appropriate equation for the unknown variable.

For example, consider a racing car that starts from rest and accelerates uniformly over a 160-meter track, crossing the finish line after 8 seconds. To find the acceleration, we first identify the known variables: the initial velocity (\(v_0 = 0\)), time (\(t = 8\) seconds), and displacement (\(\Delta x = 160\) meters). The target variable is acceleration (\(a\)). With these three known variables, we can use the third equation:

$$\Delta x = v_0t + \frac{1}{2}at^2$$

Substituting the known values gives:

$$160 = 0 \cdot 8 + \frac{1}{2}a(8^2)$$

Solving for \(a\) results in:

$$a = \frac{160 \cdot 2}{8^2} = 5 \, \text{m/s}^2$$

In the next part, to find the final velocity at the finish line, we can use the first equation:

$$v = v_0 + at$$

Substituting the known values gives:

$$v = 0 + 5 \cdot 8 = 40 \, \text{m/s}$$

By mastering these equations and understanding how to identify the relevant variables, students can effectively solve a wide range of motion problems in physics.

A car accelerates from 5 m/s to 21 m/s at a constant 3.0 m/s². How far does it travel while accelerating?

• A. 437 m
• B. 2.67 m
• C. 69.3 m
• D. 42.7 m

Answer: (C)

A moving hockey puck encounters a patch of rough ice and slides 90cm before coming to a full stop. Assuming a constant deceleration of 8 m/s², what was the puck's initial speed?

• A. 3.79 m/s
• B. 2.68 m/s
• C. 37.9 m/s
• D. 1.90 m/s

Answer: (A)

Certain rifles can fire a bullet with a speed of 970 m/s just as it leaves the muzzle (this speed is called the muzzle velocity). The muzzle is 70.0 cm long and the bullet is accelerated uniformly from rest within it. 
For how long (in ms) is the bullet in the muzzle?

• A. 144 ms
• B. 1.44 ms
• C. 0.722 ms
• D. 1440 ms

Answer: (B)

Acceleration Signs

In motion problems, understanding the relationship between acceleration, velocity, and speed is crucial. A common misconception is that positive acceleration always indicates an increase in speed. However, positive acceleration simply means that the velocity is becoming more positive, while negative acceleration indicates that the velocity is becoming more negative. This distinction is essential for correctly interpreting motion scenarios.

The formula for acceleration is given by:

$$ a = \frac{\Delta v}{\Delta t} $$

where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time. A positive value for acceleration indicates that the change in velocity is positive, while a negative value indicates a decrease in velocity.

To determine whether speed is increasing or decreasing, focus on the magnitude of the velocity, disregarding the sign. For instance, if an object moves from an initial velocity of 10 m/s to a final velocity of 30 m/s, the speed is increasing because the magnitude (30) is greater than the initial magnitude (10). Conversely, if an object moves from 30 m/s to 10 m/s, the speed is decreasing since the magnitude is reducing.

When analyzing acceleration and speed, it is important to note that:

• If the signs of acceleration and initial velocity are the same, the speed increases.
• If the signs of acceleration and initial velocity are different, the speed decreases.

For example, if an object has a positive initial velocity and positive acceleration, it speeds up. If it has a negative initial velocity and negative acceleration, it also speeds up. However, if the initial velocity is positive and the acceleration is negative, the object slows down, and similarly, if the initial velocity is negative and the acceleration is positive, the object also slows down.

To illustrate these concepts, consider a truck moving to the left at 30 m/s that comes to a stop after traveling 150 meters. The initial velocity is negative (since it moves left), and the final velocity is 0. Using the equation:

$$ v_f^2 = v_i^2 + 2a \Delta x $$

we can solve for acceleration. Substituting the known values, we find:

$$ 0 = (-30)^2 + 2a(-150) $$

which simplifies to:

$$ 0 = 900 - 300a $$

Rearranging gives:

$$ 300a = 900 $$

Thus, the acceleration is:

$$ a = 3 \, \text{m/s}^2 $$

Even though the truck is slowing down, the acceleration is positive because the direction of motion (left) is considered negative in this context.

Understanding these principles helps clarify the often confusing relationship between acceleration, velocity, and speed in motion problems.

Acceleration with Multiple Parts

When solving problems involving an object moving under constant acceleration, it's essential to approach the situation methodically. Begin by organizing the information through a diagram and identifying the five key variables: displacement (\(\Delta x\)), initial velocity (\(v_i\)), final velocity (\(v_f\)), acceleration (\(a\)), and time (\(t\)). For scenarios with multiple segments, label the variables for each segment distinctly, such as \(v_a\) for the initial velocity in segment A to B, and \(v_b\) for the final velocity in that segment.

In a typical example, consider a vehicle traveling at a constant speed of 30 meters per second. Upon noticing a road hazard, the driver takes 0.7 seconds to react before applying the brakes. This situation can be divided into two intervals: the reaction time (A to B) and the stopping time (B to C). During the reaction phase, the vehicle maintains a constant velocity, meaning the acceleration is zero. Thus, the distance traveled during this phase can be calculated using the formula:

\[ \Delta x_{A \to B} = v_{A} \cdot t_{A \to B} \]

Substituting the known values gives:

\[ \Delta x_{A \to B} = 30 \, \text{m/s} \cdot 0.7 \, \text{s} = 21 \, \text{meters} \]

Next, for the stopping phase (B to C), the initial velocity remains 30 m/s, but the final velocity is 0 m/s, with a deceleration of -6 m/s². To find the time taken to stop, we can use the equation:

\[ v_{C} = v_{B} + a \cdot t_{B \to C} \]

Rearranging this gives:

\[ 0 = 30 + (-6) \cdot t_{B \to C} \]

Solving for \(t_{B \to C}\) yields:

\[ t_{B \to C} = 5 \, \text{seconds} \]

Finally, to determine the total distance traveled from A to C, we sum the distances from both segments:

\[ \Delta x_{A \to C} = \Delta x_{A \to B} + \Delta x_{B \to C} \]

For the distance from B to C, we can use the equation:

\[ \Delta x_{B \to C} = v_{B} \cdot t_{B \to C} + \frac{1}{2} a \cdot t_{B \to C}^2 \]

Substituting the known values gives:

\[ \Delta x_{B \to C} = 30 \cdot 5 + \frac{1}{2} \cdot (-6) \cdot (5^2) = 150 - 75 = 75 \, \text{meters} \]

Thus, the total distance from A to C is:

\[ \Delta x_{A \to C} = 21 + 75 = 96 \, \text{meters} \]

This structured approach, involving clear organization and the application of kinematic equations, allows for effective problem-solving in motion scenarios involving constant acceleration.

Subway Train

In this problem, we analyze the motion of a subway train that undergoes three distinct phases: acceleration, constant speed, and deceleration. To solve for the total distance traveled by the train, we first break down the motion into segments and identify the relevant variables for each phase.

The first phase, from point A to B, involves the train accelerating from rest. Here, we denote the initial velocity \( V_A = 0 \), the acceleration \( a = 1.5 \, \text{m/s}^2 \), and the time \( t = 14 \, \text{s} \). To find the distance \( \Delta x_{AB} \), we use the equation for uniformly accelerated motion:

\[ \Delta x_{AB} = V_A t + \frac{1}{2} a t^2 \]

Substituting the known values, we find:

\[ \Delta x_{AB} = 0 \cdot 14 + \frac{1}{2} \cdot 1.5 \cdot (14)^2 = 147 \, \text{m} \]

Next, during the second phase from B to C, the train travels at a constant speed for \( t = 60 \, \text{s} \). The velocity at this stage, \( V_B \), is equal to \( V_C \) since there is no acceleration. To find \( V_B \), we use the first phase's data:

\[ V_B = V_A + a t = 0 + 1.5 \cdot 14 = 21 \, \text{m/s} \]

Now, we can calculate the distance \( \Delta x_{BC} \) using the formula for constant velocity:

\[ \Delta x_{BC} = V_B \cdot t_{BC} = 21 \cdot 60 = 1260 \, \text{m} \]

In the final phase, from C to D, the train decelerates to a stop with an acceleration of \( a = -3.5 \, \text{m/s}^2 \). The initial velocity for this segment is \( V_C = 21 \, \text{m/s} \) and the final velocity \( V_D = 0 \). We need to find the distance \( \Delta x_{CD} \) using the equation:

\[ V_D^2 = V_C^2 + 2a \Delta x_{CD} \]

Substituting the known values gives:

\[ 0 = (21)^2 + 2 \cdot (-3.5) \cdot \Delta x_{CD} \]

Rearranging and solving for \( \Delta x_{CD} \) yields:

\[ 441 = 7 \Delta x_{CD} \implies \Delta x_{CD} = \frac{441}{7} = 63 \, \text{m} \]

Finally, to find the total distance \( \Delta x_{AD} \) traveled by the train, we sum the distances from all three segments:

\[ \Delta x_{AD} = \Delta x_{AB} + \Delta x_{BC} + \Delta x_{CD} = 147 + 1260 + 63 = 1470 \, \text{m} \]

This comprehensive approach highlights the importance of organizing the problem into manageable parts and applying the appropriate kinematic equations to derive the total distance traveled by the train.

Here’s what students ask on this topic:

What are the four kinematic equations and when should each be used?

The four kinematic equations are used to solve motion problems under constant acceleration. They are:

1. $v_f=v_0+at$ - Use when you need to find final velocity.

2. $v_{f}2=v_{0}2+\; 2a\mathrm{\Delta x}$ - Use when time is not given.

3. $\mathrm{\Delta x}=v_{0}t+\; 0.5at2$ - Use when you need to find displacement.

4. $\mathrm{\Delta x}=\; 0.5(v_0+v_f)t$ - Use when acceleration is not given.

Choose the equation based on the known and unknown variables in your problem.

How do you determine the sign of acceleration in kinematics problems?

The sign of acceleration depends on the direction of the velocity change. Positive acceleration means the velocity is becoming more positive, while negative acceleration means the velocity is becoming more negative. Use the formula $a=\mathrm{\Delta v}\mathrm{\Delta t}$ to determine the sign. If both initial velocity and acceleration have the same sign, the object is speeding up. If they have opposite signs, the object is slowing down.

How do you solve a kinematics problem involving multiple parts of motion?

To solve a kinematics problem with multiple parts, follow these steps:

1. Draw a diagram and label each part of the motion.

2. List the five variables (Δx, v₀, v_f, a, t) for each part.

3. Identify known and unknown variables for each part.

4. Use the appropriate kinematic equation for each part, ensuring you account for the transition between parts.

5. Solve for the unknowns step-by-step, using the results from one part as inputs for the next.

What is the difference between velocity and speed in kinematics?

Velocity is a vector quantity that includes both magnitude and direction, while speed is a scalar quantity that only includes magnitude. For example, a velocity of 30 m/s to the right is different from a speed of 30 m/s, which does not specify direction. In kinematics, velocity is used in equations to account for direction, whereas speed is used to describe how fast an object is moving regardless of direction.

How do you find the final velocity of an object given initial velocity, acceleration, and time?

To find the final velocity (v_f) of an object given its initial velocity (v₀), acceleration (a), and time (t), use the kinematic equation:

$v_f=v_0+at$

Plug in the values for v₀, a, and t to solve for v_f. This equation is useful when you need to determine how the velocity of an object changes over time under constant acceleration.