1-Dimensional Kinematics

  • Thread starter Sonic7z
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In summary, the speeder passes a parked police car at 30.0 m/s and the police car starts from rest with a uniform acceleration of 2.44m/s2. Using the equations x=x0+v0t+1/2at^2 for the cop and x=30t for the speeder, it can be determined that the cop catches the speeder after 24.6 seconds and in that time, the speeder travels 738 meters.
  • #1
Sonic7z
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Homework Statement


A speeder passes a parked police car at 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.44m/s2.

a. How much time passes before the speeder is taken over by the police car?
b. How far does the speeder get before bing overtaken?

Homework Equations


s=Vi+at
Vf=Vi+at
 
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  • #2
Welcome to PF, sonic!
Careful with this one - there are two different kinds of motion involved. The speeder is moving at constant speed, so you need a simpler equation for the distance in that case. The equations you typed for accelerated motion are not quite right.

I suggest you make two headings, one for the speeder, one for the police car and write the appropriate equation(s) under each. Then fill in the numbers you know.

You'll need to identify what is the same for both cars and use that fact to put together a solution.

Good luck!
 
  • #3
x=x0+v0t+1/2at^2

x=position
x0=initial value of position, here call it zero
v0=initial speed, 0 for the cop, 30m/s for the speeder
a=acceleration, 2.44m/s/s for the cop, 0 for the speeder


xcop=1/2*2.44m/s/s t^2

xspeeder = 30t

the cop catches the speeder when 1.22t^2 = 30t or when t=24.6s

in this time, the speeder travels 246.s x 30m/s = 738m
 
  • #4
Looks great!
 
  • #5

Vf^2=Vi^2+2as

a. The time passed before the speeder is taken over by the police car can be calculated using the formula Vf=Vi+at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration and t is the time. In this case, the initial velocity of the police car is 0 m/s and the final velocity of the speeder is 30.0 m/s. Therefore, we can rearrange the equation to solve for t: t = (Vf-Vi)/a = (30.0 m/s - 0 m/s)/2.44 m/s^2 = 12.3 seconds. So, it would take 12.3 seconds for the police car to catch up to the speeder.

b. To calculate the distance the speeder travels before being overtaken, we can use the formula s=Vi+at, where s is the distance, Vi is the initial velocity, a is the acceleration and t is the time. In this case, the initial velocity of the speeder is 30.0 m/s and the final velocity is the same as the police car, which is 30.0 m/s. Therefore, we can rearrange the equation to solve for s: s = Vi*t + 0.5*a*t^2 = 30.0 m/s * 12.3 seconds + 0.5 * 2.44 m/s^2 * (12.3 seconds)^2 = 369.0 meters. So, the speeder would travel 369.0 meters before being overtaken by the police car.
 

1. What is 1-Dimensional Kinematics?

1-Dimensional Kinematics is the study of motion in a straight line, along a single axis. It involves analyzing the position, velocity, and acceleration of an object in motion.

2. What are the basic equations of 1-Dimensional Kinematics?

The basic equations of 1-Dimensional Kinematics are:
- Position (x) = Initial position (x0) + Velocity (v) * Time (t)
- Velocity (v) = Initial velocity (v0) + Acceleration (a) * Time (t)
- Position (x) = Initial position (x0) + Initial velocity (v0) * Time (t) + 0.5 * Acceleration (a) * Time (t)^2
- Velocity (v)^2 = Initial velocity (v0)^2 + 2 * Acceleration (a) * Change in position (Δx)

3. How do you calculate average velocity in 1-Dimensional Kinematics?

The average velocity in 1-Dimensional Kinematics is calculated by dividing the change in position (Δx) by the change in time (Δt). This can also be represented as:
Average velocity (v) = Change in position (Δx) / Change in time (Δt) = (xf - xi) / (tf - ti)

4. What is the difference between speed and velocity in 1-Dimensional Kinematics?

Speed is a measure of how fast an object is moving, while velocity is a measure of both the speed and direction of motion. In 1-Dimensional Kinematics, speed can be calculated as the absolute value of velocity, since direction is not considered.

5. How does acceleration affect an object's motion in 1-Dimensional Kinematics?

Acceleration is the rate of change of an object's velocity. If an object has a positive acceleration, it is speeding up in the positive direction. If it has a negative acceleration, it is slowing down in the positive direction. If an object has a constant acceleration, its velocity changes at a constant rate. Acceleration can also be zero, indicating that the object is either at rest or moving at a constant velocity.

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