Example 2.8 Watch Out for the Speed Limit! A car traveling at a constant speed o

Question

Example 2.8 Watch Out for the Speed Limit!

A car traveling at a constant speed of 40.0 m/s passes a trooper on a motorcycle hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch the car, accelerating at a constant rate of 3.10 m/s2. How long does it take her to overtake the car?

SOLVE IT

A pictorial representation helps clarify the sequence of events. The car is modeled as a particle under constant velocity, and the trooper is modeled as a particle under constant acceleration.

First, we write expressions for the position of each vehicle as a function of time. It is convenient to choose the position of the billboard as the origin and to set tB = 0 as the time the trooper begins moving

At that instant, the car has already traveled a distance of 40.0 m from the billboard because it has traveled at a constant speed of vx = 40.0 m/s for one second. Therefore, the initial position of the speeding car is xB = 40.0 m.

Use the following equation to give the car's position at any time t:

xcar = x B + vx, cart = 40.0 m + (40.0 m/s) t

A quick check shows that at t = 0, this expression gives the car's correct initial position when the trooper begins to move: xcar = xB = 40.0 m.

The trooper starts from rest at tB = 0 and accelerates at 3.10 m/s2 away from the origin. Use the following equation to give her position at any time t:

Explanation / Answer

1)

initially car is travelling with a constant speed v= 40 m/sec
after 1 sec distance travelled is

xB= speed*time

xB=40*1

xB=40 m

now
position of the car after "t" time is

xcar=xB+v*t

xcar=40+40*t ..............(1)

now trooper starts from the rest(u=0) and accelerates at 3.19 m/sec^2 and finally cath the car after "t" time and position of the trooper is,

xtrooper=ut+1/2*a*t^2

xtrooper=0*t+1/2*3.1*t^2

xtrooper=1/2*3.1*t^2        ..................(2)

when the trooper overtaking the car

xcar=xtrooper [from equation (1) and (2)]
  

40+40*t=1/2*3.1*t^2

===>

1.55t^2-40t-40=0

above equation is a quadratic equaton

the roots of the above are

t=[-(-40)+sqrt(40^2-4*1.55*-40)]/2*1.55        or t=[-(-40)+sqrt(40^2-4*1.55*-40)]/2*1.55

t=[40+42.988]/2*1.55       or   t=[40-42.988]/2*1.55(neglect)

t=41.494 sec ....is answer

2)

a)
initialilly car is travelling with constant spped v=37.5 m/sec
after 7sec the distance travelled is,

xo=speed*time

xo=37.5*7

xo=262.5 m

now

position of the car after "t" time is

xcar=xo+v*t

xcar=262.5+37.5*t    ..............(1)

now policeman starts from the rest and accelerates at 3.8 m/sec^2 and finally catch the car after "t" time and position of the policeman is,

xpolice=ut+1/2*a*t^2

xpolice=0*t+1/2*3.8*t^2

xpolice=1/2*3.8*t^2   ...........(2)

when the policeman overtaking the car

xcar=xpoliceman [from equation (1) and (2)]

262.5+37.5*t=1/2*3.8*t^2

===>

1.9t^2-37.5t-262.5=0

above equation is quadratic equaton

the roots of the above are

t=[-(-37.5)+sqrt(37.5^2-4*1.9*-262.5)]/2*1.9        or      t=[-(-37.5)-sqrt(37.5^2-4*1.9*-262.5)]/2*1.9

t=[37.5+58.32]/2*1.9    or    t=[37.5-58.32]/2*1.9 (neglet)

t=25.215 sec ....is answer

b)
the distance travelled by the police man is,
xpolice=1/2*3.8*t^2

=1/2*3.8*25.215^2

=1208.094 m ......is answer

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