Question

police officer at rest at side of highway notices speeder moving at 17m/s along road.when speeder passes ,officer accelerates 3 m/ssquare in pursuit.speeder do not notice until police catch up.
a)calculate speed of police car when officer catches him
b)assume police car accelerates until its moving 10km/h faster than speeder and then moves at constant velocity until catching up with speeder.how long will it take to catch speeder?

Answer 1

what's up @anitmary ?

Answer 2

not so good with all this problems.........

Answer 3

ok, so, did you get a)?

Answer 4

ya its 33m/s......but didnt get b)??

Answer 5

ok, looks like the easiest way is...

Answer 6

the concept is \[d_{cop}=d_{speeder} ryt??\]

Answer 7

yeah correct:)

Answer 8

d=192.1 m & t=11.3 s

Answer 9

for b) ?

Answer 10

not for b).........m telling how i got a)....

Answer 11

yep that's right for a)

Answer 12

10km/h = 2.78 m/s

so calculate the time and distance he covers getting up to speed
time :

Vf = Vi +at
19.78 = 0 +3t

distance:

19.78^2 = 0^2 +2*3*d

Answer 13

didnt you get time already?????????

Answer 14

next find the distance the speeder covers in that time:
(easymode) 17*t
then see how far the officer has to go still in order to catch him...

\[\huge 0 = d _{speeder} -d _{cop} -17t _{2} +19.78t _{2}\]

Answer 15

sry, equation editor is bugging, might have to restart browser.

Answer 16

sure

Answer 17

then add the time you got from part one (the acceleration phase) to the time in part two (t2)

Answer 18

t=6 s we got that ryt??

Answer 19

i didnt understand your equation......

Answer 20

which?

Answer 21

0=dspeeder-dcop-17t2......??

Answer 22

yeah I guess I wrote it weird

try this... you got a distance for the officer and the speeder in the acceleration phase right?

Answer 23

so how far is the speeder still ahead at that point?

Answer 24

didnt understand wat happened after getting t=6s???

Answer 25

so calculate the time and distance he covers getting up to speed

time :
Vf = Vi +at
19.78 = 0 +3t

distance:
Vf^2 = Vi^2 +2ad
19.78^2 = 0^2 +2*3*d

Answer 26

t= 19.78/3
d=(19.78)^2 / 6

Answer 27

meanwhile, the speeder has gone 17*(19.78/3) meters...

Answer 28

so he's ahead by a distance of 17*(19.78/3) - (19.78)^2 /6

Answer 29

agree so far?

Answer 30

ohhkk....

Answer 31

does it make sense? or is this just blowing your mind:)

Answer 32

so even wen cop's initial speed is 10 km/h faster than speeder at t=6s ,he didnt catch up with him ryt??

Answer 33

nope. not yet.. he's still behind a bit.

Answer 34

yea..ok,got it..

Answer 35

so, ready for the second bit? when he travels at a constant speed catching the rest of the way up to the speeder?

Answer 36

yes....

Answer 37

ok so
17*(19.78/3) - (19.78)^2 /6 + 17*t2 = 19.78*t2

Answer 38

t2 is the time for the second part, the part with no acceleration.

Answer 39

yes ...ok..

Answer 40

find t2, add it to the first time (acceleration phase) and that's the total time to catch up to the speeder.

Answer 41

so final time is 9.006s??? but text answer is 24s??

Answer 42

46.88 = 2.78t

Answer 43

\[d _{cop}=66.67m and d _{speeder}=17\times6.67=113.39m \]

Answer 44

just solve this for t2:
17*(19.78/3) - (19.78)^2 /6 + 17*t2 = 19.78*t2
add it to the time in the first part.

Answer 45

16.86 + 6.6

Answer 46

ok i get 23....something..
i have a doubt........during that t=6s ,,,,vi=0 m/s & vf=20m/s ryt?? or was it vi=20 m/s and vf=0 m/s???

Answer 47

Vf =19.78

Answer 48

when there is almost 46.7 m to be covered(period of constant velocity) ,,vi=20 m/s and vf=0 m/s ryt??

Answer 49

no.

Answer 50

there is no "Vf' or "Vi" for that part... both are just travelling at the same velocity throughout. driver 17 m/s officer 19.78 m/s

Answer 51

ohh...yea...i forgot