Physics Problem

Question

1.) A state trooper, observing car A traveling at an excessive speed, starts from rest 

and, with an initial, constant acceleration of 2.0 m s–2, chases the car. Car A,
initially at a distance of 100 m in front of the officer, is speeding at 24 m s–1 on the
same road and traveling away from the officer. The officer reduces his acceleration
to zero once he attains a speed of 40 m s–1 and travels at that speed thereafter. At
the instant the officer attains the speed of 40 m s–1, the driver of car A sees the
flashing lights from the police car in her rear–view mirror. Realizing that she is in
trouble, the nervous driver of car A decelerates at a constant rate, and reduces her
speed to zero as she pulls over to the shoulder. Neglect the size of the vehicles.

 

a. How long a time did it take for the officer to reach a speed of 40 m s–1?

 

b. What was the position of the driver, relative to the state trooper’s car, 15
seconds from the instant the officer started the chase?

 

c. Determine the distance the trooper traveled before he attained the speed of 40
m s–1

 

d. How far ahead (distance) was the nervous driver from the officer when she
spotted the flashing lights on the police car?

Answer 1

Joey,

 

This problem looks more complicated than it really is, simply because it asking for a lot of different things to be solved for, but lets break it down and just take things one step at a time.

 

We only need two fundamental equations of motion to solve for all of this:

 

x = x₀ + v₀·t + ½a·t²

 

and

 

v = v₀ + a·t

 

The first equation lets us solve for position (or distance) x as a function of time, if we know something about the initial velocities and times of our particles (cars).  We will use this equation to obtain information about both the trooper and the speeder.

 

The second equation lets us solve for velocities if we know starting velocities and acceleration.  We will only need to use this equation for the trooper, since he's the only one who will be changing speed during this solution.

 

First we write down what we know:  at time t = 0

 

Initial positions are:  trooper at x = 0 and speeder at x = 100m

Initial velocities:  trooper has v_t(0) = 0 and speeder has v_s(0) = 24 m/s

Initial acceleration:  trooper has a = 2m/s² and speeder has a = 0

 

The speeder will continue traveling at constant speed until the very end of the problem, so that simplifies things somewhat.  The 1st question asks how long it takes for the trooper to reach a speed of 40 m/s, so lets solve that 1st.

 

Part 1:  For the trooper use v = v₀ + a·t, v₀ = 0 and v = 40 m/s, and a = 2 m/s²

Solving for t, t = v/a = (40 m/s)/(2 m/s²) = 20 s

 

at t = 20 s, the troopers velocity = 40 m/s

 

Part2:  What is the speeder's position relative to the trooper at t = 15 s.

 

For the trooper's position x_t = ½a·t² = ½·(2 m/s²)·(15 s)² = 225 m

 

For the speeder's position x_s = x₀ + v₀·t = (100 m) + (24 m/s)·(15 s) = 460 m

 

The position of the speeder relative to the trooper at t = 15 s is then:  x_s - x_t = 460 m - 225 m = 235 m

 

The speeder was 235 m in front of the trooper at t = 15 s

 

Part3:  To determine how far the trooper had traveled when he reached a speed of 40 m/s, we can multiply his average speed  by the time (20 s) it took him to reach that speed; x = ½(40 m/s)·(20 s) = 400 m

 

x = 400 m at t = 20 s  (we could also have used x = ½a·t² = ½(2 m/s²)·(20 s)² = 400 m)

 

Part4:  For the last part we simply need to figure out how far the speeder had traveled in 20 s, the subtract the trooper distance to find out how far apart they were when she saw the flashing lights.

 

x_s = (100 m) + (24 m/s)·(20 s) = 100 m + 480 m = 580 m

 

X = x_s - x_t = 580 m - 400 m = 180 m

 

So she was 180 m in front of the trooper when she spotted his lights and began to slow down.

 

Hope this helps explain how to handle these types of problems.

 

 

 

Answer 2

Trooper initial velocity.      =0

           initial acceleration = 2mt/sec^2

 

Initial distance between them 100 Mt ( car ahead)

 

Speed of the car = 24mt/sec

 

(a)

Time for officer to reach speed 40mt/sec

V= U + at

  =2t=40

t= 20sec

 

(b)

    position of car relative to officer after 15 minutes

 

Officer moves -       S1=Ut +(1/2)a*t^2

 

                      S= (1/2)*2*15^2

                        =225 Mt

 

Car moves.-         S2=Ut

                               =24*15=360 Mt

 

Car is ahead by  (360+100-225)= 235 Mt

 

(c)

Distance traveled by Trooper before he attains speed 40mt/sec

       S=Ut+(1/2)*a*t^2

 

          = (1/2)*2*20^2=400 Mt

(d)

           Relative distance between car and officer

 

        Distance traveled by car =

           S = Ut

              24*20=480 Mt

 

   Relative distance = (480+100-400)=180mt. Car ahead by 180 mt

 

           

Answer 3

1) v=v₀+at, so since v₀=0 (officer starts from rest), a=2 m/s², you obtain:

 

40=2t, so t=20 seconds.

 

2) In this case it is better to imagine things from the driver's perspective. In its reference frame, the officer is at -100 m at t=0 with the velocity of -24 m/s and acceleration 2 m/s². In 15 seconds, the officer position is given by the following equation:

 

x=x₀+v_0xt+a_xt²/2, where x₀=-100 m, v_0x=-24 m/s, a_x=2 m/s², and t=15 seconds. Substitute everything in to find:

 

x=-100-24*15+2*15²/2=-100-360+225=-235 m; 

 

That is, officer was 235 m behind the driver.

 

3) Distance traveled in uniformly accelerated motion is:

 

d=at²/2; In our case t=20 s (see question 1), a=2m/s². 

d=2*20²/2=400 m.

 

4) Use the same approach as in question 2) to answer question 4. Only this time t=20 s. Calculation yields:

 

d=-100-24*20+2*20²/2=-100-480+400=-180 m. So the officer was 180 m behind the driver, thus driver was 180 meters ahead of the officer.