Determining the Distance Required for Overtaking a Car with Constant Acceleration

Understanding the mechanics of overtaking a car that is accelerating can be crucial for both safety and efficiency in traffic. This article explores the mathematics behind determining the distance needed for a car traveling at a constant speed to overtake another car that is accelerating from rest. The key concept involves the comparison of speeds and distances traveled over time.

Given Data and Constants

Let's define the constants and variables used in the problem:

Speed of the first car, v1 20 m/s Initial speed of the second car, u2 0 m/s (at rest) Acceleration of the second car, a2 2 m/s2 Length of each car, L 5 meters Time, t

Step 1: Determine the Time to Reach the Same Speed

The second car, initially at rest, accelerates to reach the speed of the first car. The speed of the second car at any time t can be found using the equation for velocity under constant acceleration:

v2(t) u2 a2t 0 2t 2t

Setting this equal to the speed of the first car, 20 m/s:

2t 20

t 10 seconds

Step 2: Calculate the Distance Traveled by the Second Car

The distance traveled by the second car in 10 seconds can be calculated using the formula for distance under constant acceleration:

d2 u2t frac12; a2t2 0 frac12; cdot; 2 cdot; 102 100 meters

Step 3: Calculate the Distance Traveled by the First Car

The distance traveled by the first car in 10 seconds is:

d1 v1t 20 cdot; 10 200 meters

Step 4: Calculate the Total Overtaking Distance

The total distance required for the first car to completely overtake the second car includes the distance the second car traveled and the length of both cars:

Total distance d2 2L 100 5 5 110 meters

Conclusion

The total road distance used in overtaking is 110 meters. This example illustrates the importance of careful calculation in determining safe overtaking maneuvers. By understanding these mechanics, drivers can improve traffic flow and enhance road safety.

Additional Considerations and Revisitation of the Problem

The additional calculation introduced another scenario where the relative distance and velocity are considered:

At time t0, the positions of the cars are as shown in the figure. The velocity of car 2 relative to car 1 at any time t is given by:

v 20 - 2t

For the first car to overtake the second, it needs to cover a relative distance of 10 meters with velocity v given by the equation above. The relative distance is 5 5 10 meters.

10 int;{0 to to} (20 - 2t) dt 20t - t2

to2 - 20to 10 0

Solving the quadratic equation, we get:

to 20 ± sqrt{400 - 40} / 2 10 pm; 9.49

Therefore, to 0.51 seconds or 19.49 seconds.

During time 0.51 seconds, car 2 travels the road distance:

d 20 cdot; 0.51 10.2 meters

At time t19.49 seconds, car 1 moves with velocity v1 38.98 m/s. The distance it covers is:

d1 frac12; cdot; 19.492 379.9 meters

Meanwhile, car 2 moves through distance:

d2 19.49 cdot; 20 389.8 meters

At this point, car 2 again overtakes car 1.