Cop A Catches Thief: Time to Catch & Calculation

[Sakha]

Homework Statement

A thief steals a car in front of two cops, A and B.
The thief's initial velocity is 25m/s, cop A starts his car with 3m/s², and cop B starts 10 seconds after cop A, but with 5m/s². After 200m from his initial point, the thief notices the cops and start accelerating at 1m/s²

Which cop catches the thief? At what time does he catch it?

Homework Equations

d= V_ot+1/2at²
I think that's the only required formula

The Attempt at a Solution

I tried to set an equality between the distance of the thief and the distance of cop A, but I get stuck in the math to solve for t.

(25m/s)(t)+(1/2)(m/s²)+200m=(1.5m/s²)(t+8)²
t is the time since the thief started accelerating, and that happens 8s after A started.
after doing math here I get completely stuck with the algebra.

 

[Sakha]

I tried a new equation:
(25m/s)t+(1/2)(m/s²)t²-200m = (1.5m/s²)t²
I got that t = 10.73seconds, but I don't know if its correct because the left side of the equation which is the thief's distance seems that it was always accelerating but started 200m behind cop A, so would anyone tell me what equation I could use?

 

[baba89]

I would first like to clarify the given information. The thief's initial velocity is given as 25m/s, but it is not specified in which direction the thief is traveling. I will assume that the thief is traveling in the same direction as the cops, and therefore his initial velocity is positive 25m/s.

Next, I would like to point out that the given equation, d= Vot+1/2at2, is only applicable when the acceleration is constant. In this scenario, the acceleration of the thief and the cops is not constant, as it changes at different points in time. Therefore, we cannot use this equation to solve the problem.

Instead, we can use the equation for position as a function of time, x(t) = x0 + v0t + 1/2at2, where x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time. We can use this equation to calculate the position of each person at any given time.

Let's start by setting up the equation for the thief, who starts accelerating after 200m from his initial point. His initial position, x0, is 200m, his initial velocity, v0, is 25m/s, and his acceleration, a, is 1m/s2. Therefore, the equation for the thief's position as a function of time is: x(t) = 200 + 25t + 1/2(t-200)2.

To determine when the thief is caught, we need to find the time when the thief's position is equal to the position of either cop A or B. Let's start with cop A. His initial position, x0, is 0m, his initial velocity, v0, is 3m/s, and his acceleration, a, is also 3m/s2. Therefore, the equation for cop A's position as a function of time is: x(t) = 3t + 1/2(3)t2.

To solve for the time when the thief is caught, we can set the two equations equal to each other and solve for t. This will give us the time when the thief's position is equal to cop A's position. We can then plug this time into the thief's equation to find his position at that time, which will also be cop A's position.

Similarly,