A motorist travels at a constant speed of 34.0 m/s through a school zone, exceed

Question

A motorist travels at a constant speed of 34.0 m/s through a school zone, exceeding the posted speed limit. A policeman waits 7.0 s before giving chase at an acceleration of 3.5 m/s^2.

(a) Find the time required to catch the car from the instant the car passes the policeman. (s) (Write the equations for the position of the speeder at time t and for the position of the policeman who starts 7.0 seconds later. Find the time when the two meet.)

(b) Find the distance required for the policeman to overtake the motorist. (m)

Explanation / Answer

This is a problem of cinematics

Data

Vm = 34 m/s

ap= 3.5 m/ s2

t = 7 s

where

m refers the motorized

p refers to police

Vm = X/t

Xm = Vm t

The police out 7 s later, as we measure time since leaving the moto

Xp = Vo t + ½ ap (t-7)2

At the point where is Xm = Xp

We equate the two equations

Vm t = ½ ap ( t -7 )2

We solve the equation

Vm t = ½ ap ( t2 - 2 7 t + 49)

34 t =1/2 3.5 ( t2 - 14 t + 49)

34 t = 1.75 ( t2 - 14 t + 49)

19.429 t = t2 -14 t + 49

t2 - 33.429t +49 = 0

t = (33.429 ± sqrt (33.4292 – 4 49) ) / 2

t = ( 33.429 ± sqrt (921.498) )/2 = (33.429 ± 30.356)/2

t1 = 1.5365 s

t2 = 31.892 s

Both times may be correct because they are positive, but the first encounter occasionally happen over time smaller 1.5365 s

Result 1.5365 s

Part b)

X = Vm t

X = 34 * 1.5365

Xm = 52.241 m

X = ½ 3.5 ( 1.5365 – 7)2

Xp = 52,24 m

As we can see the two distances are equal

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