Can Police Catch the Speeding Motorist?

  • Thread starter Thread starter aflowernamedu
  • Start date Start date
  • Tags Tags
    Information
Click For Summary
SUMMARY

The discussion focuses on calculating the distance at which police officers can catch a speeding motorist traveling at 108 km/hr (30 m/s). After a 2-second delay, the police car accelerates at 2.5 m/s² for 15 seconds, covering 281.25 meters. In total, the police take 47.5 seconds to reach the same point as the motorist, who travels 510 meters in that time. The intersection occurs 510 meters from the observation point where the police began their pursuit.

PREREQUISITES
  • Understanding of basic kinematics equations, specifically S = ut + 1/2*a*t²
  • Knowledge of acceleration and uniform motion concepts
  • Familiarity with speed and distance calculations
  • Ability to perform relative speed analysis
NEXT STEPS
  • Study advanced kinematics problems involving multiple objects in motion
  • Learn about relative velocity and its applications in real-world scenarios
  • Explore the effects of acceleration on pursuit scenarios in physics
  • Investigate the use of calculus in motion analysis for more complex situations
USEFUL FOR

Students studying physics, law enforcement professionals interested in pursuit dynamics, and anyone involved in traffic safety analysis.

aflowernamedu
Messages
3
Reaction score
0
How would I solve this?

2 police officers are standing on a road. A motorist comes at 108km/hr(30m/s) The police take 2 s to assess the situation. Then their patrol car accelerates at 2.5m/s^2 for 15s after the 2s and then continues at uniform speed. How far from the observation point does the intersection take place? Assume the motorist maintains a constant speed of 108km/hr?
 
Physics news on Phys.org
Yes you do, assuming the question means the police react as the bike passes parallel to them.

using:

S = ut + 1/2*a*t2

You find that the police travel 281.25m in the 15 seconds they accelerate, plus the 2 seconds they waited, so in 17s they have traveled 281.25m and accelerated to a constant velocity of 37.5m/s

In this time the motorbike has traveled 17*30 = 510m so the police after 17s are a distance of 228.75m behind the bike

The relative speed between the two is 7.5m/s

So how long does it take to make up 228.75m at 7.5 m/s?

228.75 / 7.5 = 30.5s

plus the original 17s

30.5 + 17 = 47.5s until the police are level with the biker.
 
Thank you ,but I still am not sure of the answer to "how far from the observation point did the intersection take place?"
 

Similar threads

Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
1
Views
1K
  • · Replies 18 ·
Replies
18
Views
8K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K