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

[anglum]

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

 

[anglum]

what should my next steps be?? please help me

 

[anglum]

can someone point me in thee right direction so i can start tryin to solve this the correct way

 

[anglum]

learningphysics can you help me

 

[learningphysics]

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?

 

[anglum]

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?

 

[learningphysics]

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?

 

[anglum]

i am gettin screwed up becuz the speeds are in km/h and the time is in seconds GRRR i suck

 

[anglum]

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?

 

[learningphysics]

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?

 

[anglum]

yes that is the acceleration

 

[learningphysics]

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...

 

[anglum]

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

 

[anglum]

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 ?

 

[learningphysics]

yes. that looks right... at the end remember to add 18.888s to whatever you get here to get total time.

 

[anglum]

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) ?

 

[anglum]

cuz if i do that the distances are different

 

[learningphysics]

cuz if i do that the distances are different

Take the speeder's velocity times the total time.

 

[anglum]

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

 

[anglum]

so again the distances and times are equal for the two cars correct?

 

[iskar]

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.

 

[anglum]

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?

 

[learningphysics]

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.

 

[anglum]

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?

 

[iskar]

sorry (170*170) - 0 = 2 * d * 9 ... then solve for distance and use this to calculate the time.

 

[anglum]

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?

 

[learningphysics]

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

 

[anglum]

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

 

[learningphysics]

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.

 

[anglum]

ok so when solving that i get

t = 24.81927711 seconds? is that correct?

 

[anglum]

but what is that the time of? i am so lost right now

 

[learningphysics]

ok so when solving that i get

t = 24.81927711 seconds? is that correct?

yes.

 

[anglum]

what is that 24.8 the time of?\

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

 

[anglum]

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

 

[learningphysics]

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?