Straight-line Kinematics problem

In summary, the conversation is about a physics problem involving two cars moving at 90 km/h, with one suddenly applying brakes and the other following after a delay. The question is about the minimum separation between the two cars to avoid collision. The answer given in the book is different from the one calculated using a specific formula. The conversation also discusses the difference between using a slide-rule and a calculator for calculations. Ultimately, it is concluded that the answer in the book is incorrect.
  • #1
LD_90
11
0
I found this problem in the text "Elementary Physics Classical and Modern" by Weidner and Sells.

Two automobiles are both moving at 90 km/h in the same direction, one directly behind the other. The driver of the lead car suddenly applies his brakes, decelerating at 7.5 m/s^2. The other driver applies his own brakes after a delay of 0.40 s, and slows down at a rate of 6.0 m/s^2. (His tires are worn.) If there is to be no collision, what is the minimum separation between the cars at the instant the lead car's brakes are applied?

The answer given in the back of the book is "greater than 30 5/6 m or about 30.9 m." I'm wondering if this anwer is correct.

Let the lead car be car A: vo=25m/s , v=0 , a=-7.5m/s^2
using the formula v^2=vo^2+2a(x-xo) , car A stops after 41.7m, car B after 52.8m+10m(for the delay). I can get a close anwer by using a different approach. Why doen't the first method give the correct answer? :confused:
 
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  • #2
LD_90 said:
Let the lead car be car A: vo=25m/s , v=0 , a=-7.5m/s^2
using the formula v^2=vo^2+2a(x-xo) , car A stops after 41.7m, car B after 52.8m+10m(for the delay). I can get a close anwer by using a different approach. Why doen't the first method give the correct answer? :confused:
This method seems correct to me. I get a minimum lead of 20.4m to avoid a collision. What other method did you use?
 
  • #3
Thanks Doc Al. That's the answer that I got too. Since it did not match the back of the book answer, I took the difference in stopping times, (plus the delay) and multiplied it by the initial velocity. This gives about 30.8m. This gets a result closer to the book answer, but it doesn't seem right to me. Oh by the way, the book was published pre-calculator in 1973 with the original in the mid sixties, so I'm not sure how far slide-rule results tend to be from calculator results.
 
  • #4
LD_90 said:
Since it did not match the back of the book answer, I took the difference in stopping times, (plus the delay) and multiplied it by the initial velocity.
That makes no sense to me.
Oh by the way, the book was published pre-calculator in 1973 with the original in the mid sixties, so I'm not sure how far slide-rule results tend to be from calculator results.
Come on, things weren't that bad back then! :rolleyes:
 
  • #5
Well I guess the answer in the book is wrong. Thanks for the help.
 

1. What is straight-line kinematics?

Straight-line kinematics is a branch of physics that deals with the motion of objects in a straight line without considering the causes of motion.

2. What types of problems can be solved using straight-line kinematics?

Straight-line kinematics can be used to solve problems involving the motion of objects in a straight line, such as calculating position, velocity, and acceleration.

3. What is the difference between distance and displacement in a straight-line kinematics problem?

Distance is the total length traveled by an object, while displacement is the shortest distance between the initial and final position of an object.

4. How do you calculate velocity in a straight-line kinematics problem?

Velocity is calculated by dividing the change in displacement by the change in time. In other words, velocity = displacement/time.

5. What is the significance of the slope on a distance-time graph in a straight-line kinematics problem?

The slope on a distance-time graph represents the velocity of the object. A steeper slope indicates a higher velocity, while a flatter slope indicates a lower velocity.

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