Solve Car Chase Question: 110 km/h and 170 km/h

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A speeder traveling at 110 km/h is pursued by a cop car starting from rest and accelerating at 9 km/h² until it reaches a maximum speed of 170 km/h. The time taken for the cop car to reach 170 km/h is calculated to be approximately 18.88 seconds, during which the speeder travels about 0.577 km. To determine the total time for the cop to catch the speeder, the distances traveled by both vehicles are set equal, leading to a total catch-up time of approximately 26.79 seconds. The discussion also touches on similar problems involving different speeds and accelerations, emphasizing the importance of setting distances equal for both vehicles to solve for time. The conversation highlights the complexities of converting units and applying kinematic equations effectively.
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Homework Statement


A speeder traveling 110 km/h races past a cop car. The cop car is at rest and pursues at acceleration of 9km/h until it reaches its maximum speed of 170km/h until it catches up with the speeder. How long does it take the cop car to catch the speeder answer in units of seconds

The Attempt at a Solution



i know the distance the 2 travel is equal

so for the speeder d = 110/t

and for the cop car i should use what formula so that i can relate them to d=d

so that 11/t = the distance formula for the cop car
 
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what should my next steps be?? please help me
 
can someone point me in thee right direction so i can start tryin to solve this the correct way
 
learningphysics can you help me
 
anglum said:
learningphysics can you help me

I recommend first find out the distance between the two cars, when the cop car has reached 170km/hr...

How long does it take the cop car to reach 170km/h? Over what distance does this happen?

How long has the speeder traveled in this time?
 
ok so the time for the cop car to get to 170 km/h is

170/9 = 18.8888 is that in seconds then?

so then in 18.88 seconds the otehr car would have gone .577133 km?
 
Last edited:
anglum said:
ok the time it takes the cop to reach the 170 km/h is 18.888 seconds... but I am not sure how to calculate how long it takess him to catch the other car please help

What distance has the cop traveled in 18.888 seconds? What distance has the speeder driven in 18.888s?
 
i am gettin screwed up becuz the speeds are in km/h and the time is in seconds GRRR i suck
 
ok so am i right in my sayin it takes the cop 18.88 seconds to reach 170 km/h

and then i converted 170 to km/s aand got .0472 km/s
and i converted 110 km/h to km/s and got .03055 km/s

then the distance the speeder went was .57713 km
and the distance for the cop is what?
 
  • #10
anglum said:
i am gettin screwed up becuz the speeds are in km/h and the time is in seconds GRRR i suck

Is the acceleration 9km/h per second? What exactly is the acceleration given?
 
  • #11
yes that is the acceleration
 
  • #12
anglum said:
yes that is the acceleration

For the cop, you'd use:

d = v1*t + (1/2)at^2. here v1 = 0. convert 9km/hr per second into km/s^2...
 
  • #13
the distance the cop travels then is.445568 km after 18.88 seconds

so now i knwo they are .131562 km apart at that time

so now what do i do to solve for the time of the cop car to catch the speeder
 
  • #14
so i then get this

cop ---- d+.131562 = .0472km/s (t)
speeder ---- d = .03055km/s (t)

so then i can say that .0472t - .131562 = .03055t ?
 
  • #15
yes. that looks right... at the end remember to add 18.888s to whatever you get here to get total time.
 
  • #16
so the total time it takes to catch teh speeder is 26.7896 seconds

now to find the total distance each car traveled i simply take

d = .03055km/s (26.7896) or d +.131562 = .0472km/s (26.7896) ?
 
  • #17
cuz if i do that the distances are different
 
  • #18
anglum said:
cuz if i do that the distances are different

Take the speeder's velocity times the total time.
 
  • #19
one more question regarding car chases learning if u don't mind

speeder goin 30.9 m/s

cop accelerates at 2.49m/s squared

how long until he catches speeder
 
  • #20
so again the distances and times are equal for the two cars correct?
 
  • #21
Like you said initially all you need is a distance formula for the cop car...

final velocity squared - initial velocity squared = 2 * distance * acceleration

so for the cop: 170 - 0 = 2 * d * 9

just solve for distance and then use that in your initial equation for the speeder.
 
  • #22
my other problem is ...

speeder goes by cop at 30.9m/s

cop accelerates at 2.49m/s squared

how logn till he catches speeder

again the distance the 2 travel and the time they travel is the same correct?

so what equation should i solve since i don't know how far ahead the speeder is this time?
 
  • #23
anglum said:
one more question regarding car chases learning if u don't mind

speeder goin 30.9 m/s

cop accelerates at 2.49m/s squared

how long until he catches speeder

same type of thing as before... The position of the speeder is 30.9t.

What is the position of the cop car?

set the two positions equal. Solve for time.
 
  • #24
the position of the cop car is .5(2.49)tsquared

if i set them to equal i don't know how to solve for t?
 
  • #25
sorry (170*170) - 0 = 2 * d * 9 ... then solve for distance and use this to calculate the time.
 
  • #26
the position of the cop car is .5(2.49)tsquared

if i set them to equal i don't know how to solve for t?

or is that wrong for the position of cop car?
 
  • #27
anglum said:
the position of the cop car is .5(2.49)tsquared

if i set them to equal i don't know how to solve for t?

0.5(2.49)t^2 = 30.9t

1.245t^2 = 30.9t

1.245t^2 - 30.9t = 0

t(1.245t - 30.9) = 0
 
  • #28
ok but once it gets to that point how do i solve for t from there

t(1.245t -30.9) = 0

wont anything i solve equal 0 ... grr i need to stop doing theese late at nite when my brain is fried
 
  • #29
anglum said:
ok but once it gets to that point how do i solve for t from there

t(1.245t -30.9) = 0

wont anything i solve equal 0 ... grr i need to stop doing theese late at nite when my brain is fried

When you have ABC... = 0, that means A = 0 or B = 0 or C = 0 etc...

t(1.245t -30.9) = 0

implies:

t = 0 or 1.245t - 30.9 = 0

the t = 0 is at the start... it's not the solution you need. solve: 1.245t - 30.9 = 0.
 
  • #30
ok so when solving that i get

t = 24.81927711 seconds? is that correct?
 
  • #31
but what is that the time of? i am so lost right now
 
  • #32
anglum said:
ok so when solving that i get

t = 24.81927711 seconds? is that correct?

yes.
 
  • #33
what is that 24.8 the time of?\

and what steps do i take next i am so lost on this problem
 
  • #34
o nevermind wow its late... that's the time it takes to catch up.. .i forgot what we were solving for jeeezzzz I am a moron
 
  • #35
anglum said:
what is that 24.8 the time of?\

and what steps do i take next i am so lost on this problem

That's the time the one car catches up with the other... they start at the same time... on car quickly goes past the other with constant speed... the other car accelerates, and eventually goes faster than the other car... so it catches up..

Are there more parts?
 
  • #36
ok so can i move onto the next problem i have...

a ball is thrown at an angle of 50degrees at 17m/s ,,,, how long does it take to reach its maximum height??
 
  • #37
anglum said:
ok so can i move onto the next problem i have...

a ball is thrown at an angle of 50degrees at 17m/s ,,,, how long does it take to reach its maximum height??

What is its vertical velocity at its maximum height?
 
  • #38
im not sure what that is... these are my toughest problems
 
  • #39
anglum said:
im not sure what that is... these are my toughest problems

At the max. height, the vertical velocity is 0. What is the initial vertical velocity?
 
  • #40
the initial velocity is 17m/s but its at a 50 degree angle so what would its vertical velocity be
 
  • #41
anglum said:
the initial velocity is 17m/s but its at a 50 degree angle so what would its vertical velocity be

17sin(50)
 
  • #42
ok so the vertical velocity is 13.022m/s? so then i can plug that into the

Vf= Vi - AT

and solve for T and that is the time to reach its vertical max?
 
  • #43
this question has 2 more parts

this ball is shot by a person at a height of 2.626m and goes thru a hoop 3.048m high

what is the distance of the shot... and how long does it take to reach the hoop?

again its shot at 17m/s at angle of 50 degrees
 
  • #44
What is the vertical displacement of the ball?

Use d = (vsin(theta))*t + (1/2)(-g)t^2 to find the time...

Then find the horizontal distance using vcos(theta)*t
 
  • #45
what the heck is vsin? theta? -g?

the vertical displacement is .422 meters correct?
 
  • #46
anglum said:
what the heck is vsin? theta? -g?

the vertical displacement is .422 meters correct?

Yes. Try to use the equation d = v1*t + (1/2)at^2 in the vertical direction... that's the equation I gave you v = 17. theta = 50. g = 9.8
 
  • #47
but i don't know what d or t are so how do i solve for that?
 
  • #48
anglum said:
but i don't know what d or t are so how do i solve for that?

d=0.422. solve for t.
 
  • #49
but that won't be the answer for how long it takes the ball to travel the horizontal distance to the hoop... the hoop is x away from the person... how can i solve how long it takes to reach the hoop without knowing how far the hoop is from teh person?
 
  • #50
anglum said:
but that won't be the answer for how long it takes the ball to travel the horizontal distance to the hoop...

it is the same time.

the hoop is x away from the person... how can i solve how long it takes to reach the hoop without knowing how far the hoop is from teh person?

It is the same time. 17cos(50)*t gives the horizontal distance.
 

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