'Police Car Catching Up' problem in kinematics

[Ineedhelpwithphysics]

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

???

 

[Ineedhelpwithphysics]

do i add one because after one second???

 

[erobz]

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?

 

[Ineedhelpwithphysics]

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

 

[erobz]

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 $t+1$.

 

[erobz]

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?

 

[Ineedhelpwithphysics]

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

 

[erobz]

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.

 

[kuruman]

Imagine that each vehicle has its own clock registering time t_c for the car and t_p for the police. Then the equations that you wrote should be
Cars displacement = 30*t_c +1/2(0)t_c²
police displacement = 0*t_p + 1/2(6)t_p²

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

What equation relates the time on the car clock to the time on the police clock?

 

[Ineedhelpwithphysics]

Imagine that each vehicle has its own clock registering time t_c for the car and t_p for the police. Then the equations that you wrote should be
Cars displacement = 30*t_c +1/2(0)t_c²
police displacement = 0*t_p + 1/2(6)t_p²

The equation for the car predicts that when the clock starts at t_c = 0, the car is at x = 0.
The equation for the police predicts that when the clock starts at t_p = 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?

 

[erobz]

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 $t_c$.

 

[Lnewqban]

The displacement equation?

 

[haruspex]

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?

 

[brotherbobby]

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.

 

[brotherbobby]

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.

 

[Lnewqban]

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.

 

[PeroK]

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 $t =0$.