When and Where Will the Second Car Overtake the First? A Comprehensive Analysis

In this article, we will analyze a classic problem in physics and solve it step by step, breaking it down into two main parts - the acceleration phase and the constant speed phase. By understanding the principles of acceleration and velocity, we can determine the exact moment and distance when the second car overtakes the first car.

Introduction to the Problem

The problem involves two cars that start from the same initial position. Car 1 accelerates at a rate of 1 m/s2 for 10 seconds, while Car 2 accelerates at a rate of 0.8 m/s2 for 20 seconds. Both cars then continue to move at their final speeds. The question we aim to answer is: when will the second car overtake the first and at what distance?

Acceleration Phase Analysis

Car 1 (1 m/s2 Acceleration)

tAcceleration: ( a_1 1 , text{m/s}^2 ) tTime of Acceleration: ( t_1 10 , text{s} ) tFinal Speed of Car 1: tt[ v_1 a_1 cdot t_1 1 , text{m/s}^2 cdot 10 , text{s} 10 , text{m/s} ] tDistance Traveled by Car 1 During Acceleration: tt[ d_1 frac{1}{2} a_1 t_1^2 frac{1}{2} cdot 1 , text{m/s}^2 cdot (10 , text{s})^2 50 , text{m} ]

Car 2 (0.8 m/s2 Acceleration)

tAcceleration: ( a_2 0.8 , text{m/s}^2 ) tTime of Acceleration: ( t_2 20 , text{s} ) tFinal Speed of Car 2: tt[ v_2 a_2 cdot t_2 0.8 , text{m/s}^2 cdot 20 , text{s} 16 , text{m/s} ] tDistance Traveled by Car 2 During Acceleration: tt[ d_2 frac{1}{2} a_2 t_2^2 frac{1}{2} cdot 0.8 , text{m/s}^2 cdot (20 , text{s})^2 160 , text{m} ]

Summary After Acceleration

After 10 seconds, Car 1 has traveled 50 meters and is moving at 10 m/s, while Car 2 has traveled 160 meters and is moving at 16 m/s. Therefore, Car 2 is ahead of Car 1 by 110 meters after 10 seconds.

Constant Speed Phase Analysis

Once the acceleration phase is over, both cars continue to move at their final speeds. Let's denote the time after the initial 10 seconds as ( t ). We can express the distances traveled by each car after the initial 10 seconds as follows:

Distance Traveled by Car 1

[ d_{text{Car 1}} 50 10t ]

Distance Traveled by Car 2

[ d_{text{Car 2}} 160 16t ]

Setting the distances equal to determine when Car 2 overtakes Car 1:

[ 50 10t 160 16t ]

Now, solve for ( t ):

[ 50 - 160 16t - 10t ] [ -110 6t ] [ t approx 18.33 , text{s} ]

Total Time and Distance at Overtaking

The total time from start until Car 2 overtakes Car 1 is:

[ text{Total time} 10 , text{s} 18.33 , text{s} 28.33 , text{s} ]

The distance at which Car 2 overtakes Car 1 can be calculated using the distance formula for either car:

[ d_{text{Car 1}} 50 , text{m} 10t approx 50 , text{m} 10 cdot 18.33 , text{s} 233.33 , text{m} ]

Hence, Car 2 overtakes Car 1 at approximately 28.33 seconds and 233.33 meters.

Understanding the mechanics of acceleration, velocity, and distance can help solve real-world problems in transportation, engineering, and physics. This analysis demonstrates how these principles can be applied to determine the overtaking point of two vehicles on a straight path.