Kinematics Crash Simulation Question

Click For Summary
In a crash simulation, the stopping distance of a car traveling at double the speed is determined using the equation V^2 = Vi^2 + 2ad, which shows that the stopping distance quadruples when speed is doubled. The initial misunderstanding arose from using an average speed equation, which is less applicable in this context. The correct approach involves recognizing the relationship between speed, acceleration, and distance under constant deceleration. Additionally, the time taken to stop can be calculated using V = Vi + at, which was not clearly understood by the original poster. Ultimately, the correct answer to the question is D, indicating both the stopping time doubles and the stopping distance quadruples.
tobya93
Messages
3
Reaction score
0
1. In a crash simulation, a car traveling at x m/s can stop at a distance d m with a maximum deceleration. If the car is traveling at 2x m/s:

A. Stopping time is doubled
B. Stopping distance is doubled
C. Stopping distance is quadrupled
D. Both A and C


Ok, so I thought this was a relatively easy question. However, I managed to butcher it. My problem is, I do not understand which equation to use. To find if the distance doubled or quadrupled I used the equation: d = ((Vi + V)/2) * t. Since the car will be stopping the V will be zero and you simply get d = (Vi/2) * t. If the velocity is doubled, distance is doubled. However, the book used the equation V^2 = Vi^2 + 2ad. Thereby getting, stopping distance quadrupled.

Also for time they used V = Vi + at.
Answer is D.

Once again, I am lost to know which equation to use. Since both seemed to work..
Thanks for your help!
 
Physics news on Phys.org
Hello Toby, and welcome to PF.
Pity you didn't use the template. Any good reason for that ? Because now I don't know what material (equations) you have available to tackle this.
Also, variable names, given/known data are a bit messy now.

Your d = ((Vi + V)/2) * t is not very useful, since you don't know t (or did you assume t is the same for x m/s and 2x m/s ?). It is also very seldom you can use average speed, so better stay away from this equation if possible (by happenstance it is in fact correct here)

1.) Are you familiar with d = x * t + 1/2 * a * t2 for motion under constant acceleration(/deceleration) ?
Usually written as ## S = v_0\, t + {1\over 2}\, at^2##

2.)
for time they used V = Vi + at
gives the impression you don't recognize this. Is that really unfamiliar ?

3.) Do you recognize
V^2 = Vi^2 + 2ad
? In another way: do you know about kinetic energy and the work done by a braking force that acts over a distance d ?
 
I get a notification "Arka420 has just replied ..." time stamp 20140623 15:22:15 +0000 but don't get to see it posted. Some good spirit got crossed at the considerable giveaway therein? But an identical notification must have been mailed to Toby at he same instant, isn't it?

If spiritual activity was indeed called for, leave something less telepathic for mere earthlings, please. Is Toby being helped meaningfully now (or rather, since originally posting, two weeks ago)?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
3
Views
938
  • · Replies 7 ·
Replies
7
Views
2K
Replies
18
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
18
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K