A policeman is traveling due north with a velocity of18.0 m/s when a fleeing ban

Question

1. A policeman is traveling due north with a velocity of18.0 m/s when a fleeing bank robber zooms by with a velocity of42.0 m/s due north. After a reaction time of 0.8 s the policemanbegins to pursue the robber with an acceleration of 5.0m/s2.

1. What distance will the policeman travel in catching therobber?
2. What is the policeman’s speed, in mph, when hecatches up to the robber?

I attempted to solve this problem but I keptgetting lost. I started by using the equation: V0x*a*t=Vx 18m/s*5.0m/s2.8=22m/s (Im not sure if I went aboutthis the right way) This is where I got stuck because I tried to use thex=V0x*t+1/2*ax*t2 and my answercame out completly wrong which then threw of part b of thisquestion. Can anyone help me out?

1. What distance will the policeman travel in catching therobber?
2. What is the policeman’s speed, in mph, when hecatches up to the robber?

I attempted to solve this problem but I keptgetting lost. I started by using the equation: V0x*a*t=Vx 18m/s*5.0m/s2.8=22m/s (Im not sure if I went aboutthis the right way) This is where I got stuck because I tried to use thex=V0x*t+1/2*ax*t2 and my answercame out completly wrong which then threw of part b of thisquestion. Can anyone help me out? 18m/s*5.0m/s2.8=22m/s (Im not sure if I went aboutthis the right way) This is where I got stuck because I tried to use thex=V0x*t+1/2*ax*t2 and my answercame out completly wrong which then threw of part b of thisquestion. Can anyone help me out?

Explanation / Answer

1.   dp:t= = distance the policemanhas traveled at time ''

2.   vp:t= = velocity of policeman attime ''

3.   ap:t= = acceleration ofpoliceman at time ''

4.   dr:t= = distance the robber hastraveled at time ''

5.   vr:t= = velocity of robber attime ''

.

06.   At: t = 0.0

07.      dp:t=0 = 0.0

08.      vp:t=0 = 18.0m/s

09.      ap:t=0 = 0.0

10.      dr:t=0 = 0.0

11.      vr:t=0 = 42.0m/s

.

12.   At: t = 0.8

13.     dp:t=0.8 =dp:t=0 + (vp:t=0)*(t) = 0 + (18.0 m/s)*(0.8s) = 14.4 meters

14.      vp:t=0.8 = 18.0m/s

15.      ap:t=0.8 = 5.0m/s2

16.      dr:t=0.8 =dr:t=0 + (vr:t=0)*(t) = 0 + (42.0 m/s)*(0.8s) = 33.6 meters

17.      vr:t=0.8 = 42.0m/s

.

18.   In the next few steps, solve for the time ittakes the policeman’s distance to equal the robber’sdistance.

19.   At: t =         ;  when: dp = dr

20.      (dp:t=0.8) +(vp:t=0.8)*() +(ap:t=0.8)*(2) = (dr:t=0.8) +(vr:t=0.8)*() +(ar:t=0.8)*(2)

21.      (14.4 m) + (18.0 m/s)*()+ (5.0 m/s2)*(2) = (33.6 m) + (42.0m/s)*() + (0.0 m/s2)*(2)

22.      In ‘21’, combinelike terms and arrange the terms in quadratic form.

23.      (5.0)*(2) +(-24.0)*() + (-19.2) = 0

24.      Solve ‘23’ usingquadratic formula: a=5, b=-24, c=-19.2

25.      = (24 ±30.98387)/(10)      ; must bepositive, therefore, = 5.498387 seconds

.

26.   Calculate the total distance the policeman hastraveled (check against robber’s distance).

27.      dp:t= =(dp:t=0.8) + (vp:t=0.8)*() +(ap:t=0.8)*(2)

28.      dp:t= = (14.4m) + (18.0 m/s)*(5.498387) + (5.0m/s2)*[5.498387)2] = 264.532240 meters

29.      dr:t= = (33.6m) + (42.0 m/s)*(5.498387) + (0.0m/s2)*[5.498387)2] = 264.532240 meters

.

30.   Calculate the velocity of the policeman at t=

31.      vp:t= =(vp:t=0.8) + (ap:t=0.8)*() = (18.0 m/s)+ (5.0 m/s2)*( 5.498387) = 45.491933 meters/second

32.      Convert the result in‘31’ to miles per hour.

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