Q2

A motorcycle is traveling at a speed of 72 $km/h$. It will come to a stop with a deceleration of 4 $m/s^2$ at a traffic light. What is the minimum safe distance required for the motorcycle to stop safely from the point where the brakes are applied?

The Problem: Stopping Distance

We have a motorcycle moving at a certain speed and then decelerating to a stop. We want to find the minimum distance it covers during this braking process. This is what we call the stopping distance.

1. Identify What We Know (Given Information)

• Initial Speed ($v_0$): The speed of the motorcycle before it applies the brakes.
  • $v_0=72\text{ km/h}$
• Final Speed ($v$): The speed of the motorcycle after it comes to a stop.
  • $v=0\text{ m/s}$ (It stops!)
• Deceleration ($a$): The rate at which the motorcycle's speed decreases.
  • $a=4{\text{ m/s}}^2$. Since it's deceleration (slowing down), we'll treat this as a negative acceleration in our equations: $a=-4{\text{ m/s}}^2$. (The acceleration vector points opposite to the velocity vector).

2. Identify What We Want to Find

• Stopping Distance ($\mathrm{\Delta }x$): The minimum distance required for the motorcycle to stop.

3. Crucial First Step: Unit Conversion!

Physics problems always require consistent units. Notice our speed is in kilometers per hour (km/h) but our acceleration is in meters per second squared (m/s²). We must convert the initial speed to meters per second (m/s).

Here's how we do it:

• 1 km = 1000 m
• 1 hour = 3600 s

$v_0=72\text{ km/h}\times \frac{1000\text{ m}}{1\text{ km}}\times \frac{1\text{ h}}{3600\text{ s}}$

$v_0=\frac{72\times 1000}{3600}\text{ m/s}$

$v_0=\frac{72000}{3600}\text{ m/s}$

$v_0=20\text{ m/s}$

Now all our units are consistent (meters and seconds)!

4. Choose the Right Kinematic Equation

We're looking for distance ($\mathrm{\Delta }x$), and we know initial speed ($v_0$), final speed ($v$), and acceleration ($a$). We don't have time ($t$) and we don't want to calculate it if we don't have to.

The best kinematic equation for this scenario is:

$v^2=v_0^2+2a\mathrm{\Delta }x$

This equation directly relates initial velocity, final velocity, acceleration, and displacement ($\mathrm{\Delta }x$).

5. Plug in the Values and Solve!

Now, let's substitute our known values into the equation:

$(0\text{ m/s}{)}^2=(20\text{ m/s}{)}^2+2(-4{\text{ m/s}}^2)\mathrm{\Delta }x$

$0=400{\text{ m}}^2/{\text{s}}^2-8{\text{ m/s}}^2\cdot \mathrm{\Delta }x$

Now, we need to isolate $\mathrm{\Delta }x$:

Add $8{\text{ m/s}}^2\cdot \mathrm{\Delta }x$ to both sides:
$8{\text{ m/s}}^2\cdot \mathrm{\Delta }x=400{\text{ m}}^2/{\text{s}}^2$

Divide both sides by $8{\text{ m/s}}^2$:
$\mathrm{\Delta }x=\frac{400{\text{ m}}^2/{\text{s}}^2}{8{\text{ m/s}}^2}$

$\mathrm{\Delta }x=50\text{ m}$

6. The Answer and What It Means

The minimum safe distance required for the motorcycle to stop is 50 meters.

Why "minimum safe distance"? This calculation assumes ideal braking conditions – the brakes are applied instantly, the deceleration is constant, and there are no external factors like reaction time or varying road conditions. In reality, a driver's reaction time and less-than-perfect braking can significantly increase the actual stopping distance.

Q3

"Two cars start from the same point. Car A accelerates from 0 m/s to 100 m/s over 20 seconds, while Car B travels at a constant speed of 60 m/s. Let's evaluate the following statements:

1. The displacement of Car A from 0 to 20 seconds is 1000 meters.
2. At t = 12 seconds, Car A and Car B are at the same point.
3. The average speed of Car A from 0 to 20 seconds is less than the average speed of Car B over the same time interval.
4. If the car movement continues until t = 25 seconds, Car A will overtake Car B.

Excellent analysis work, class! This problem tests our ability to apply kinematics uniformly and carefully consider the details of motion for each object. Let's break down each statement.

First, let's establish the equations of motion for both cars. Both cars start from the same point, so we can set their initial position $x_0=0$.

Car A Analysis:

• Initial velocity $v_A=0\text{ m/s}$
• Final velocity (at $t=20\text{ s}$) $v_{A,f}=100\text{ m/s}$
• Time interval $\mathrm{\Delta }t=20\text{ s}$

Since Car A accelerates uniformly, we can find its acceleration:
$a_A=\frac{v_{A,f}-v_A}{\mathrm{\Delta }t}=\frac{100\text{ m/s}-0\text{ m/s}}{20\text{ s}}=5{\text{ m/s}}^2$

Now we can write the kinematic equations for Car A for $0\le t\le 20\text{ s}$:

• Velocity: $v_A(t)=v_A+a_{A}t=0+5t=5t\text{ m/s}$
• Position: $x_A(t)=v_{A}t+\frac{1}{2}{a}_A{t}^2=0\cdot t+\frac{1}{2}(5)t^2=2.5t^2\text{ m}$

Car B Analysis:

• Constant speed $v_B=60\text{ m/s}$

The kinematic equation for Car B:

• Position: $x_B(t)=v_{B}t=60t\text{ m}$

Now, let's evaluate each statement:

Statement 1: The displacement of Car A from 0 to 20 seconds is 1000 meters.

We can use Car A's position equation at $t=20\text{ s}$:
$x_A(20)=2.5\times (20{)}^2=2.5\times 400=1000\text{ m}$
Alternatively, using the average velocity for constant acceleration: $\mathrm{\Delta }x=\overline{v}t=\frac{v_0+v_f}{2}t=\frac{0+100}{2}\times 20=50\times 20=1000\text{ m}$.

Conclusion for Statement 1: TRUE.

Statement 2: At t = 12 seconds, Car A and Car B are at the same point.

Let's find the position of each car at $t=12\text{ s}$:

• Position of Car A at $t=12\text{ s}$:
  $x_A(12)=2.5\times (12{)}^2=2.5\times 144=360\text{ m}$
• Position of Car B at $t=12\text{ s}$:
  $x_B(12)=60\times 12=720\text{ m}$

Since $360\text{ m}\ne 720\text{ m}$, the cars are not at the same point. Car B is ahead of Car A.

Conclusion for Statement 2: FALSE.

Statement 3: The average speed of Car A from 0 to 20 seconds is less than the average speed of Car B over the same time interval.

• Average speed of Car A (0 to 20 s):
  For constant acceleration, average speed is ${\overline{v}}_A=\frac{v_A+v_{A,f}}{2}=\frac{0\text{ m/s}+100\text{ m/s}}{2}=50\text{ m/s}$.
  (Or total distance / total time: $1000\text{ m}/20\text{ s}=50\text{ m/s}$).
• Average speed of Car B (0 to 20 s):
  Since Car B travels at a constant speed, its average speed is simply $v_B=60\text{ m/s}$.

Comparing the average speeds: $50\text{ m/s}<60\text{ m/s}$.

Conclusion for Statement 3: TRUE.

Statement 4: If the car movement continues until t = 25 seconds, Car A will overtake Car B.

This statement requires careful interpretation. "Car A accelerates from 0 m/s to 100 m/s over 20 seconds" describes the motion during the first 20 seconds. It typically implies that after 20 seconds, unless stated otherwise, Car A would maintain the velocity it achieved at the end of that acceleration period.

So, let's establish Car A's motion for $t>20\text{ s}$:

• At $t=20\text{ s}$: $x_A(20)=1000\text{ m}$ and $v_A(20)=100\text{ m/s}$.
• For $t>20\text{ s}$, Car A travels at a constant velocity of $100\text{ m/s}$.
  The position of Car A for $t>20\text{ s}$ is:
  $x_A(t)=x_A(20)+v_A(20)\times (t-20)=1000+100(t-20)$

Now, let's find the positions at $t=25\text{ s}$:

• Position of Car A at $t=25\text{ s}$:
  $x_A(25)=1000+100(25-20)=1000+100(5)=1000+500=1500\text{ m}$
• Position of Car B at $t=25\text{ s}$:
  $x_B(25)=60\times 25=1500\text{ m}$

At $t=25\text{ s}$, both cars are at the same position, $1500\text{ m}$. "Overtake" means one car passes the other and is then ahead. Since they are at the exact same point, Car A has caught up to Car B, but has not yet overtaken it. For Car A to overtake Car B, it would need to be ahead of Car B.

(Note: If we had assumed Car A's acceleration of $5{\text{ m/s}}^2$ continued indefinitely past 20 seconds, Car A would have been at $x_A(25)=2.5(25{)}^2=1562.5\text{ m}$, putting it ahead. However, the phrasing "accelerates … over 20 seconds" usually defines the event and its duration, not an ongoing acceleration. The default after such an event, without further instruction, is constant velocity at the achieved speed.)

Conclusion for Statement 4: FALSE.

Final Summary:

• Statement 1: TRUE
• Statement 2: FALSE
• Statement 3: TRUE
• Statement 4: FALSE

Qso

Determine the masses m₁ and m₃ in the three-mass, two-pulley system shown, given that θ₁ = 30°, θ₂ = 60°, and m₂ = 0.5M.

Given Information:

• Angle ${\theta }_1={30}^{\circ }$
• Angle ${\theta }_2={60}^{\circ }$
• Mass $m_2=0.5m$ (where 'm' is some reference mass)
• The system is in equilibrium.

Step 1 & 2: Free-Body Diagrams and Forces

Let's look at each mass individually. Remember, for ideal pulleys and ropes, the tension in a continuous rope section is constant.

• For mass $m_1$:
  • Downward force: Gravity, $m_{1}g$
  • Upward force: Tension, $T_1$
  • Since the rope connecting $m_1$ to $m_2$ goes over a pulley, the tension in that part of the rope is also $T_1$.
• For mass $m_3$:
  • Downward force: Gravity, $m_{3}g$
  • Upward force: Tension, $T_2$
  • Similarly, the tension in the rope connecting $m_3$ to $m_2$ is $T_2$.
• For mass $m_2$:
  • Downward force: Gravity, $m_{2}g$
  • Upward-left force: Tension, $T_1$, acting at an angle ${\theta }_1={30}^{\circ }$ from the vertical.
  • Upward-right force: Tension, $T_2$, acting at an angle ${\theta }_2={60}^{\circ }$ from the vertical.

Step 3: Apply Newton's First Law ($\mathrm{\Sigma }\overrightarrow{F}=0$)

Since the system is in equilibrium, the net force on each mass is zero.

A. For mass $m_1$:
The forces are only in the vertical direction.
$\mathrm{\Sigma }{F}_y=T_1-m_{1}g=0$
$⟹T_1=m_{1}g$ (Equation 1)

B. For mass $m_3$:
The forces are only in the vertical direction.
$\mathrm{\Sigma }{F}_y=T_2-m_{3}g=0$
$⟹T_2=m_{3}g$ (Equation 2)

C. For mass $m_2$:
This is where we need to resolve forces into their components. Let's use a standard Cartesian coordinate system, with positive x to the right and positive y upwards.

• Vertical (y) components:

  • The vertical component of $T_1$ is $T_1\cos ({\theta }_1)$ (upwards).
  • The vertical component of $T_2$ is $T_2\cos ({\theta }_2)$ (upwards).
  • The gravitational force $m_{2}g$ is downwards.

  So, $\mathrm{\Sigma }{F}_y=(T_1\cos ({\theta }_1))+(T_2\cos ({\theta }_2))-m_{2}g=0$
  $⟹T_1\cos ({\theta }_1)+T_2\cos ({\theta }_2)=m_{2}g$ (Equation 3)

• Horizontal (x) components:

  • The horizontal component of $T_1$ is $T_1\sin ({\theta }_1)$ (to the left, so negative).
  • The horizontal component of $T_2$ is $T_2\sin ({\theta }_2)$ (to the right, so positive).

  So, $\mathrm{\Sigma }{F}_x=(T_2\sin ({\theta }_2))-(T_1\sin ({\theta }_1))=0$
  $⟹T_2\sin ({\theta }_2)=T_1\sin ({\theta }_1)$ (Equation 4)

Step 4: Solve the System of Equations

Now we have a set of equations and known values. Let's substitute and simplify.

Substitute Equation 1 ($T_1=m_{1}g$) and Equation 2 ($T_2=m_{3}g$) into Equation 4:
$(m_{3}g)\sin ({\theta }_2)=(m_{1}g)\sin ({\theta }_1)$
We can cancel 'g' from both sides:
$m_3\sin ({\theta }_2)=m_1\sin ({\theta }_1)$

Plug in the given angles:
$m_3\sin ({60}^{\circ })=m_1\sin ({30}^{\circ })$
$m_3\left(\frac{\sqrt{3}}{2}\right)=m_1\left(\frac{1}{2}\right)$
Multiply both sides by 2:
$\sqrt{3}{m}_3=m_1$ (Equation 5)

Now, substitute Equation 1 and Equation 2 into Equation 3:
$(m_{1}g)\cos ({\theta }_1)+(m_{3}g)\cos ({\theta }_2)=m_{2}g$
Again, cancel 'g' from all terms:
$m_1\cos ({\theta }_1)+m_3\cos ({\theta }_2)=m_2$

Plug in the given angles and $m_2=0.5m$:
$m_1\cos ({30}^{\circ })+m_3\cos ({60}^{\circ })=0.5m$
$m_1\left(\frac{\sqrt{3}}{2}\right)+m_3\left(\frac{1}{2}\right)=0.5m$
Multiply both sides by 2:
$\sqrt{3}{m}_1+m_3=m$ (Equation 6)

Now we have a system of two equations with two unknowns ($m_1$ and $m_3$):

1. $m_1=\sqrt{3}{m}_3$
2. $\sqrt{3}{m}_1+m_3=m$

Substitute Equation 5 into Equation 6:
$\sqrt{3}(\sqrt{3}{m}_3)+m_3=m$
$3m_3+m_3=m$
$4m_3=m$
$⟹m_3=\frac{m}{4}$

Now that we have $m_3$, we can find $m_1$ using Equation 5:
$m_1=\sqrt{3}{m}_3$
$m_1=\sqrt{3}\left(\frac{m}{4}\right)$
$⟹m_1=\frac{\sqrt{3}}{4}m$

Final Answer:
Therefore, the masses are:

• $m_1=\frac{\sqrt{3}}{4}m$
• $m_3=\frac{1}{4}m$