1D Kinematics: Distance between 2 cars

Click For Summary
SUMMARY

The discussion focuses on calculating the minimum initial distance (d) between two cars to avoid a collision during braking. Car A and Car B initially travel at 60 ft/s, with Car B decelerating at 8 ft/s² and Car A at 20 ft/s² after a 0.75-second reaction time. The calculations involve using kinematic equations to determine the positions of both cars over time, ultimately leading to the conclusion that the minimum distance d must account for the distance traveled during the reaction time and the subsequent deceleration phases of both vehicles.

PREREQUISITES
  • Understanding of 1D kinematics and motion equations
  • Familiarity with the concepts of acceleration and deceleration
  • Ability to solve quadratic equations
  • Knowledge of the significance of reaction time in motion problems
NEXT STEPS
  • Study kinematic equations in detail, particularly xf=xi+vo*t+(1/2)a*t²
  • Learn about relative motion and how to analyze two objects moving towards each other
  • Explore advanced topics in dynamics, including collision avoidance strategies
  • Practice problems involving reaction time and braking distances in physics
USEFUL FOR

This discussion is beneficial for students studying physics, particularly those focusing on kinematics, as well as educators seeking to clarify concepts related to motion and collision avoidance.

Nikstykal
Messages
29
Reaction score
1

Homework Statement


Car A is traveling a distance d behind Car B. Initically both cars are traveling at the same speed of 60 ft/s. Suddenly Car B applies the brakes, causing Car B to decelerate at 8ft/s2. It takes the driver of Car A 0.75 seconds to react, and when she applies her brakes Car A decelerates at 20ft/s2.

What is the initial minimum distance d between the cars so as to avoid a collision?

Homework Equations


xf=xi+vo*t+(1/2)a*t2
vf=vo+a*t

The Attempt at a Solution


FIrst thing I did was solve for xf.b=0+60(0.75)+(-8*.5*0.752) = 42.75ft
Secondly, I solved for xf.a=0+60(0.75) = 45ft

So I know after 0.75 seconds, the distance d = c+2.25. Now I need to solve for c.

I know that A will continually get closer to B until their velocities are equal again so I then set vf=vo+a*t equal to each other for each car (once I found final velocity of B after 0.75 seconds) and solved for t, which was 0.5 seconds.

After that I solved xf=xi+vo*t+(1/2)a*t2 for each car again, getting Δxa=27.5, Δxb=26. 27.5-26 = c, which i then substituted in d = c + 2.25 to get the answer.

I got the right answer, but I am really unsure about the process. I thought my way through it but it was really roundabout. Is there a more logical process to solving this?
 
Physics news on Phys.org
Nikstykal said:
I got the right answer, but I am really unsure about the process. I thought my way through it but it was really roundabout. Is there a more logical process to solving this?

pl. explain your doubts about your calculation .
 
Which part are you unsure about? Your reasoning seems solid.
 
Thank you for the quick reply. I feel like I was just solving things randomly, maybe it is just because this stuff is new and I don;t know exactly what I am doing yet. I wasn't confident in my approach until I got the right answer.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K