Help with math word problem

Question

A highway patrol car clocks a speeder moving a constant velocity of 18 meters per second. The patrol car starts from a still position and accelerates at a rate of 5 meters per second squared. Using s(t), a function of distance find how many seconds it will take the patrol car to intercept the speeder. Let a= acceleration in meters per second squared, t=time in seconds and v=velocity in meters per second SPEEDERS DISTANCE; s(t) = vt. PATROL CARS DISTANCE: s(t)=1/2 at squared

Answer 1

Assume that the patrol car starts moving at the same time and place the speeder passes it. So, both vehicles have the same starting point. So finding when and where the patrol car overtakes the speeder is a matter of finding when/where both cars cover the same distance.

 

Let's come up with the distance equations as a function of time for the patrol and speeder.

 

Speeder:

 

V_o = 18 m/s  

V(t) = 18, speed is constant 

S(t) = ∫(18)dt = 18t + S_o -> S_o = 0, starting point is set to zero.

S(t) = 18t

 

Patrol Car:

 

A(t) = 5 m/s² , 

V(t) = ∫5dt = 5t + V_o -> 5t , V_o= 0, patrol car starts at rest.

S(t) = ∫5t = (5/2)t²

 

The patrol will overtake the speeder when the distances covered match,

 

(5/2)t² = 18t -> (5/2)t = 18 -> t = 36/5, t = 7.2 s. 

They will also intercept at 18 m/s * 7.2s -> 129.6 m.

Answer 2

SPEEDERS DISTANCE: s(t) = v t  

                                      v= 18 m/s

 

PATROL CARS DISTANCE: s(t)=1/2 a t²

                                      a= 5 m/s²

patrol car take the same distance as the speeder to intercept it, so

 

 

 v t = 1/2 a t²

18 t = 1/2 * 5 * t²

  18  =1/2 * 5 t

      t = 36/5

      t = 7.2

 

it will take the patrol car 7.2 seconds to intercept the speeder