To tackle this problem, we need to analyze the motion of the car in two distinct phases: the acceleration phase and the deceleration phase. We can use the equations of motion to derive the relationships between the distances covered, the accelerations, and the time intervals. Let’s break it down step by step.
Phase 1: Acceleration
In the first phase, the car starts from rest and accelerates with a constant acceleration \( a_1 \) for a time interval \( t_1 \). The distance covered during this phase, denoted as \( s_1 \), can be calculated using the equation of motion:
- Distance formula: \( s_1 = u t_1 + \frac{1}{2} a_1 t_1^2 \)
Since the initial velocity \( u \) is zero (the car starts from rest), the equation simplifies to:
- Distance covered: \( s_1 = \frac{1}{2} a_1 t_1^2 \)
Phase 2: Deceleration
In the second phase, the car decelerates with a constant retardation \( a_2 \) until it comes to rest. The time taken for this phase is \( t_2 \), and the distance covered is \( s_2 \). The equation of motion for this phase is:
- Distance formula: \( s_2 = u' t_2 - \frac{1}{2} a_2 t_2^2 \)
Here, \( u' \) is the final velocity at the end of the first phase, which can be calculated as:
- Final velocity: \( u' = a_1 t_1 \)
Substituting \( u' \) into the distance formula for the second phase gives us:
- Distance covered: \( s_2 = (a_1 t_1) t_2 - \frac{1}{2} a_2 t_2^2 \)
Relating the Distances and Time Intervals
Now we have expressions for both distances:
- For \( s_1 \): \( s_1 = \frac{1}{2} a_1 t_1^2 \)
- For \( s_2 \): \( s_2 = a_1 t_1 t_2 - \frac{1}{2} a_2 t_2^2 \)
To find the ratio \( \frac{s_2}{s_1} \), we can substitute the expressions we derived:
- Ratio: \( \frac{s_2}{s_1} = \frac{a_1 t_1 t_2 - \frac{1}{2} a_2 t_2^2}{\frac{1}{2} a_1 t_1^2} \)
Now, simplifying this ratio:
- Multiply both the numerator and denominator by 2:
- Resulting in \( \frac{2(a_1 t_1 t_2 - \frac{1}{2} a_2 t_2^2)}{a_1 t_1^2} \)
This simplifies to:
- Final ratio: \( \frac{s_2}{s_1} = \frac{2t_2}{t_1} - \frac{a_2 t_2^2}{a_1 t_1^2} \)
Establishing the Relationships
Now, we can also relate the accelerations and time intervals. From the equations of motion, we know:
- Using \( v = u + at \): For the first phase, \( v = a_1 t_1 \) and for the second phase, \( 0 = a_1 t_1 - a_2 t_2 \). This gives us:
- Relationship: \( a_1 t_1 = a_2 t_2 \) or \( \frac{a_1}{a_2} = \frac{t_2}{t_1} \)
Combining these relationships, we find:
- Final conclusion: \( \frac{s_2}{s_1} = \frac{a_1}{a_2} = \frac{t_2}{t_1} \)
This shows that the ratios of the distances, accelerations, and time intervals are indeed equal, confirming the relationship you were asked to demonstrate. Each phase of motion is interconnected, illustrating the beauty of kinematics in physics.
