'Police Car Catching Up' problem in kinematics

In summary, the "Police Car Catching Up" problem in kinematics involves analyzing the motion of two vehicles: a police car and a speeding vehicle. The problem typically requires calculating when and where the police car, starting from rest and accelerating at a constant rate, will catch up to the vehicle that is moving at a constant speed. To solve it, one must set up equations of motion for both vehicles, equate their distances traveled, and solve for time and distance. This problem illustrates concepts of relative motion, acceleration, and the application of kinematic equations.
  • #1
Ineedhelpwithphysics
43
7
Homework Statement
A car traveling at 30 m/s passes a cop and keeps traveling at
that speed. One second after the car passes, the cop starts accelerating from rest at 6 m/s2 until he catches up
to the car. How many seconds does it take for the cop to catch up to the car, and how many meters must the
cop travel to catch up to the car?
Relevant Equations
delta x = v initial * t + ';l1/2(a)(t)^2
Why is it 10.92 seconds and not 10

Cars displacement = 30*t +1/2(0)t^2
police displacement = 0*t + 1/2(6)(t)^2

30t = 3t^2
t = 10 seconds

???
 
Physics news on Phys.org
  • #2
do i add one because after one second???
 
  • #3
Ineedhelpwithphysics said:
do i add one because after one second???
The cop accelerates from rest after 1s has passed. So if the time the car travels before the cop catches it is ##t##, How long (time) did the cop travel?
 
  • #4
erobz said:
The cop accelerates from rest after 1s has passed. So if the time the car travels before the cop catches it is ##t##, How long (time) did the cop travel?
t+1 so its 11 not 10.92
 
  • #5
Ineedhelpwithphysics said:
t+1 so its 11 not 10.92
No, ##t## is the time when the cop catches the car. The cop has not even been accelerating for ##t##, let alone ##t+1##.
 
  • #6
Let ##t## equal 1 second. What should be the cops displacement at that time? That is how far should your equation should say the cop has moved 1 s after the car has passed it. Does your equation for the cops displacement at ##t## equals 1 second agree?
 
  • #7
erobz said:
Let ##t## equal 1 second. What should be the cops displacement at that time? That is how far should your equation should say the cop has moved 1 s after the car has passed it. Does your equation for the cops displacement at ##t## equals 1 second agree?
oh displacement is -30 from the cop to the car if i choose the car to be zero, and its 30 meters from the cop to the car if i choose the cop to be 0
 
  • #8
Ineedhelpwithphysics said:
oh displacement is -30 from the cop to the car if i choose the car to be zero, and its 30 meters from the cop to the car if i choose the cop to be 0
You could offset the displacement of the car to be 30 m ahead of the cop at ##t=0##, leaving ##t## the time the cop car accelerates to catch the car. Then you add the second the cop sat still, to find the total from the instant the car passed the cop, to the time the cop catches the car. That would work, but it's not what I was hinting at. I was hoping you would just adjust the amount of time the cop accelerates at ##a##, in comparison to the total time to catch the car.
 
  • #9
Imagine that each vehicle has its own clock registering time tc for the car and tp for the police. Then the equations that you wrote should be
Cars displacement = 30*tc +1/2(0)tc2
police displacement = 0*tp + 1/2(6)tp2

The equation for the car predicts that when the clock starts at tc = 0, the car is at x = 0.
The equation for the police predicts that when the clock starts at tp = 0, the police is at x = 0.

What equation relates the time on the car clock to the time on the police clock?
 
  • #10
kuruman said:
Imagine that each vehicle has its own clock registering time tc for the car and tp for the police. Then the equations that you wrote should be
Cars displacement = 30*tc +1/2(0)tc2
police displacement = 0*tp + 1/2(6)tp2

The equation for the car predicts that when the clock starts at tc = 0, the car is at x = 0.
The equation for the police predicts that when the clock starts at tp = 0, the police is at x = 0.

What equation relates the time on the car clock to the time on the police clock?
The displacement equation?
 
  • #11
Ineedhelpwithphysics said:
The displacement equation?
No, he is asking you what the equation relates ##t_p## and ##t_c##. It's not a kinematic equation, can you write ##t_p## in terms of ##t_c##.
 
  • #12
Ineedhelpwithphysics said:
The displacement equation?
Police-speeding cars.jpg
 
  • #13
Lnewqban said:
View attachment 339559
I couldn't make sense of the text box.
Do you mean, when the police car catches up, the areas under the lines that are to the left of that time are equal?
 
  • #14
Ineedhelpwithphysics said:
Homework Statement: A car traveling at 30 m/s passes a cop and keeps traveling at
that speed. One second after the car passes, the cop starts accelerating from rest at 6 m/s2 until he catches up
to the car. How many seconds does it take for the cop to catch up to the car, and how many meters must the
cop travel to catch up to the car?
Relevant Equations: delta x = v initial * t + ';l1/2(a)(t)^2

Why is it 10.92 seconds and not 10

Cars displacement = 30*t +1/2(0)t^2
police displacement = 0*t + 1/2(6)(t)^2

30t = 3t^2
t = 10 seconds

???
Please learn ##\rm{LaTeX}## typing in order to enable a better reading of your workings.
 
  • #15
Ineedhelpwithphysics said:
do i add one because after one second???

Could you understand where your error is @Ineedhelpwithphysics ? I hope the following helps.

Let me write for you the kinematic equations for uniform acceleration ##a_0## where all terms have their usual meanings. Just note that ##x(t_0)=x_0## and that ##v(t_0)=v_0##. Here they are :
\begin{align}
&v(t)=v_0+a_0(t-t_0)\\
&x=x_0+v_0(t-t_0)+\dfrac{1}{2}a_0(t-t_0)^2\\
&v^2(x)=v^2_0+2a_0(x-x_0)\\
\end{align}
Note that these equations are the most general. The time ##t_0>0##, which is the case for your problem.
 
Last edited:
  • Like
Likes PeroK
  • #16
haruspex said:
I couldn't make sense of the text box.
Do you mean, when the police car catches up, the areas under the lines that are to the left of that time are equal?
Sorry, my English is poor.
Yes, that is the idea I tried to bring to the attention of @Ineedhelpimbadatphys .
The distance and the time between the two points at which both cars meet (pass and catch up) are unique for both, IMHO.
I can modify the diagram later following your advice.
 
  • #17
Perhaps the simplest approach is to take ##t =0## when the cop car starts. Then you have a single time variable ##t##. The motion of the car then has an initial displacement of ##30m## at ##t =0##.
 
  • Like
Likes jbriggs444 and erobz

FAQ: 'Police Car Catching Up' problem in kinematics

What is the 'Police Car Catching Up' problem in kinematics?

The 'Police Car Catching Up' problem is a classic kinematics problem where a police car starts from rest or a certain speed and attempts to catch up to a speeding vehicle that has a head start. The problem involves calculating the time and distance it takes for the police car to reach the speeding vehicle using principles of relative motion, acceleration, and velocity.

How do you set up the equations for the 'Police Car Catching Up' problem?

To set up the equations, you need to define the initial conditions and variables. Let the initial distance between the police car and the speeding vehicle be \( d \). Assume the speeding vehicle travels at a constant speed \( v_s \) and the police car accelerates from rest with acceleration \( a \). The position of the speeding vehicle at time \( t \) is \( x_s = v_s t \). The position of the police car at time \( t \) is \( x_p = \frac{1}{2} a t^2 \). The police car catches up when \( x_p = d + x_s \).

How do you solve for the time it takes for the police car to catch up?

To find the time \( t \) it takes for the police car to catch up, set the position equations equal to each other: \( \frac{1}{2} a t^2 = d + v_s t \). This is a quadratic equation in the form \( \frac{1}{2} a t^2 - v_s t - d = 0 \). Solve for \( t \) using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = \frac{1}{2} a \), \( b = -v_s \), and \( c = -d \).

What assumptions are made in the 'Police Car Catching Up' problem?

Several assumptions are typically made: the road is straight and level, the speeding vehicle maintains a constant speed, the police car accelerates uniformly, and external factors like friction and air resistance are negligible. These assumptions simplify the problem to focus on the kinematic equations of motion.

How does the initial speed of the police car affect the problem?

If the police car starts with an initial speed \( v_p \) instead of from rest, the position equation for the police car changes to \( x_p = v_p t + \frac{1}{2} a t^2 \). The new equation to solve becomes \( v_p t + \frac{1}{2} a t^2 = d + v_s t \

Similar threads

Calculating the Time for a Car Chase: Kinematics Question
Replies
4
Views
3K
Solving a Puzzling Physics Problem: ##15 \pm 3\sqrt{15}##
Replies
17
Views
2K
Oblique soccer shot calculation task
2
Replies
38
Views
3K
8.01 MIT OCW PS1.4: Throw and Catch (Kinematics)
Replies
4
Views
832
Confused about kinematic equations
Replies
20
Views
757
Pushing a stalled car out of an intersection
Replies
1
Views
2K
Simple Work Problem Disagreeing with Kinematics
Replies
4
Views
1K
Will the Cars Meet During the Chase?
Replies
4
Views
2K
How Late Did the Man Arrive to Catch the Train?
2
Replies
36
Views
3K
Calculate the distance traveled by 2 vehicles(reaching the same point)
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top