One-Dimentional Kinematics Question

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In summary, the conversation discusses a problem involving two cars on a straight highway, where one car is traveling east with a constant velocity and the other car is traveling west with a different velocity. The first car is speeding up with a given acceleration while the second car is slowing down with a different given acceleration. The conversation also mentions using the equation X = Xi + ViT + 1/2 AT^2 to solve for the position of both cars at a given time. The person asking for help is having trouble with finding the position of the second car and is unsure if they are doing something wrong or if there is an error in the textbook's answer.
  • #1
joeisconfused
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Homework Statement


"Two cars drive on a straight highway. At time t = 0, car 1 passes mile marker 0 traveling due east with a speed of 20.0 m/s. At the same time, car 2 is 1.0 km east of mile marker 0 traveling at 30.0 m/s due west. Car 1 is speeding up with an acceleration of magnitude 2.5 m/s^2, and car 2 is slowing down with an acceleration of magnitude 3.2 m/s^2. Write x-versus-t equations of motion for both cars"


Homework Equations



The equation I used was X = Xi + ViT + 1/2 AT^2

Sorry, I'm bad at typing these equations without subscripts or superscripts.

The Attempt at a Solution



I got the equation right for Car 1. However, when I went to solve for Car two, it did not match with the answer key. I plugged in all of the relevant information into the equation:

X = 1000m + (-30 m/s)t + 1/2(3.2 m/s^2)t^2 = 1000m + (-30 m/s)t + (1.6 m/s^2)t^2

Since Car 2 is going towards the 0 marker, the velocity should be zero. However, in order for there to be deceleration, accerleration and velocity should have opposite signs. Since velocity is negative, I made acceleration positive.

However, this doesn't match the textbook answer, which is X = 1000m - (-30 m/s)t - (1.6 m/s)t^2

Am I doing something wrong?

Thanks in advance for taking the time to answer my question!
 
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  • #2
How about starting with two separate equations (as the problem suggests), one for each car. Fill in the blanks

x1 =

x2 =

Then it should be clearer what you're doing.
 
  • #3
Hm...the textbook answer looks wrong; inserting t=1, we would get X = 1028.4m > 1000m, which is clearly not possible given that the car is heading towards the 0 mark.
 
  • #4
Basically, the first one is simple enough to do without any trouble, and it's the second car that I'm having trouble with. To clear things up:

What I did for Car 1:

x1 = xi + vit + 1/2 at2
x1 = 0 m + (20 m/s)t + 1/2(2.5 m/s2)t2
x1= (20 m/s)t + (1.25 m/s2)t2

This answer matched with the textbook answer, so I didn't have any trouble for the first one. I simply plugged in the givens into the equation.

What I did for Car 2:

x2 = 1000m + (-30 m/s)t + 1/2(3.2 m/s2)t2
x2= 1000m + (-30 m/s)t + (1.6 m/s2)t2

My reasoning: Since Car 2 is going towards the 0 marker, the velocity should be zero. However, in order for there to be deceleration, accerleration and velocity should have opposite signs. Since velocity is negative, I made acceleration positive.

textbook answer is X2 = 1000m - (-30 m/s)t - (1.6 m/s)t2

Am I doing things incorrectly or is this a textbook error?

Edit: Just saw the above post. Thanks for the response!
 

FAQ: One-Dimentional Kinematics Question

What is one-dimensional kinematics?

One-dimensional kinematics is the study of motion along a single axis or line. It involves analyzing the position, velocity, and acceleration of an object in one direction.

How is displacement different from distance?

Displacement is the change in position of an object, measured in a specific direction. Distance, on the other hand, is the total length of the path traveled by the object. Displacement takes into account direction, while distance does not.

What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both the speed and direction of an object's motion. In other words, velocity includes the direction of movement, while speed does not.

How is acceleration calculated?

Acceleration is calculated by dividing the change in an object's velocity by the time it took for that change to occur. The formula for acceleration is a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

What is the difference between average and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time it took for that displacement to occur. Instantaneous velocity is the velocity of an object at a specific moment in time. It is calculated by finding the slope of the object's position vs. time graph at that specific time.

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