Two Kinematics Problems [SOLVED]

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In summary, the first problem requires setting two equations equal to each other, one for each player, and then solving for time and speed. The second problem also requires setting two equations equal to each other, one for each vehicle, and then solving for the time.
  • #1
vertciel
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[SOLVED] Two Kinematics Problems

Hello everyone,

I am having some trouble with the following two kinematic problems and I would appreciate any help or hints.

Thank you!

---

1. Two rugby players are running towards each other. They are 37 m apart. If one is accelerating from rest at [tex] 0.5 m/s^2 [/tex] and the other was already moving at 3.1 m/s and maintains her speed,

a) how long before they crunch together?
b) how fast was the accelerating player going?

My Work :

I am unsure of this, but I think I would need to set two equations equal to each other: each equation representative of each of the two rugby players. However, I do not know which equations I would need to set equal to each other.

2. A police car stopped at a set of lights has a speeder pass it at 27.78 m/s. If the police car can accelerate at [tex] 3.6 m/s^2 [/tex],

a) how long does it take to catch the speeder?

My Work :

I think that the concept is the same; I would need to set one equation for the police car and one for the speeder equal to each other.
 
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  • #2
vertciel said:
Hello everyone,

I am having some trouble with the following two kinematic problems and I would appreciate any help or hints.

Thank you!

---

1. Two rugby players are running towards each other. They are 37 m apart. If one is accelerating from rest at 0.5 m/s^2 and the other was already moving at 3.1 m/s and maintains her speed,

a) how long before they crunch together?
b) how fast was the accelerating player going?

My Work :

I am unsure of this, but I think I would need to set two equations equal to each other: each equation representative of each of the two rugby players. However, I do not know which equations I would need to set equal to each other.

You are correct. You would use a system. Try applying [tex] v^2 = {v_0}^2 +2a\Delta x[/tex] to both players.
 
  • #3
The sum of the distance the two travel has to equal 37m for the first one. If you need the time then then the best equation to use is [itex] s= ut +1/2at^2[/itex].

Again for 2.), once the speeder crosses the line and the police car sets off they both have to travel the same distance for the police car to have caught up.
 
  • #4
Thank you for your replies.

@Foxjwill, so are you recommending me to use [tex] (v_2)^2 = (v_1)^2 + 2ad [/tex]?

@Kurdt, are you recommending me to use [tex] displacement = (v_1)t + \frac{at^2}{t} [/tex] ? Also, I tried to represent the sum of the distance of the two as such but I did not get the right answer:

[tex] (v_1)t + \frac{at^2}{t} + v_1t + \frac{at^2}{t} = 37 [/tex]

Meanwhile, I will continue to work on the second problem.

Thanks again.
 
  • #5
Ok well its [itex]\frac{at^2}{2} [/itex] just for clarification. If you tell me the result it might help because its a tricky quadratic equation to solve.
 
  • #6
Thanks for your reply.

The answers are:

1. a) 7.5 s
b) 3.8 m/s

2. a) 15s
 
  • #7
vertciel said:
Thank you for your replies.

@Foxjwill, so are you recommending me to use [tex] (v_2)^2 = (v_1)^2 + 2ad [/tex]?

You can possibly use that to simplify the quadratics, but for some reason I had thought you were solving for distance. Oops.
 
  • #8
vertciel said:
Thanks for your reply.

The answers are:

1. a) 7.5 s
b) 3.8 m/s

Well for part 1 a) I got a similar result. I used 7.45 and it was fairly accurate. If you use 7.45 with [itex]s = ut [/itex] you will get, 23.095m and if you use it with [itex] 1/2 at^2 [/itex] you obtain 13.825 which when summed gives 36.92.

All I can imagine is that you haven't reported the answer to the sufficient accuracy required.
 
  • #9
Thank you for your replies, foxjwill and kurdt. I got the answers!
 

What are two common types of kinematics problems?

The two common types of kinematics problems are projectile motion and uniform circular motion. Projectile motion involves the motion of an object in a parabolic path, while uniform circular motion involves the motion of an object in a circular path at a constant speed.

How do you solve a projectile motion problem?

To solve a projectile motion problem, you first need to break the motion into horizontal and vertical components. Then, use equations of motion, such as the kinematic equations, to find the final position, final velocity, and time of flight.

What is the difference between speed and velocity?

Speed is a scalar quantity that measures the rate of change of distance, while velocity is a vector quantity that measures the rate of change of displacement. This means that velocity takes into account the direction of motion, while speed does not.

What is the difference between average velocity and instantaneous velocity?

Average velocity is the average rate of change of displacement over a specific time interval, while instantaneous velocity is the rate of change of displacement at a specific instant in time. This means that average velocity considers the overall motion, while instantaneous velocity focuses on a specific moment during the motion.

How is acceleration related to velocity?

Acceleration is the rate of change of velocity. This means that if an object's velocity is changing, then it is accelerating. If an object's velocity is constant, then it has zero acceleration. Additionally, the direction of acceleration is always in the same direction as the change in velocity.

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