A thief is noticed by a policeman from a distance of 200 m the thief starts running and the policeman chases him

Question

A thief is noticed by a policeman from a distance of 200 m the thief starts running and the policeman chases him the thief and the policemen run at the rate of 10kmph and 11 kmph respectively what is the distance between them after 6 minutes

 

let distance be x

 

D              S        T

x             10       6

x+200     11        6

 

 

speed convert in msph 10*5/18,  11*5/18

 

t=x/25/6

t=x+200/55/18

 

equation should be 

x/25/6+(x+200)/55/18=12

why i am taking 12 minutes because t1+t2

 

but answer goes wrong so please explain me where i did wrong 

 

Answer 1

Hi Amey,

 

The basic equation from Kinematics that you want to use here is

 

x = x₀ + v₀t + (1/2)at²

 

Where x₀ is the initial position, v₀ is the initial velocity and a is the acceleration. For both the thief and the cop, we assume that they run at a constant velocity, and so there is no acceleration, so a = 0. 

 

The thief and the cop each have their own position equation. Let the variables associated with the thief have a subscript "1", and left the variables associated with the cop have a subscript "2". Let's assume that the thief and the cop are running along the positive x axis. Let's assume that the thief has an initial position of 200 m and that the cop is initial on the origin. Let the thief run at a speed of v₁ = 10 km/h and the cop run at a constant speed of v₂ = 10 km/h.

 

The thief's equation is then

x₁ = 200 +  v₁t

 

The cop's equation is

 

x₂ = 0 + v₂t, or x₂ = v₂t

 

We want the difference of the thief's and cop's position at t = 6 min.

 

x₁ - x₂ = (200 + v₁t) - v₂t = 200 + (v₁ - v₂)t

 

We now have the equation that we need. But, we have to express everything in basic units according to the SI system. We want positions in meters, velocities in meters/second and times in seconds.

 

v₁ = (10 km/h)(1000 m / km)(1h / 3600 s) = 2.78 m/s

 

v₂ = (11 km/h)(1000 m / km)(1h / 3600 s) = 3.06 m/s

 

t = (6 min)(60 s / min) = 360 s

 

Thus,

x₁ - x₂ = 200 + (2.78 - 3.05)(360) = 99.2 m

 

At t = 6 min, the thief is 99.2 meters ahead of the policeman.  

Answer 2

You used kilometers in your equation (11 and 10). At the same time, you used meters (200). You need to convert one to the other first before you approach the problem.