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

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Homework Help Overview

The discussion revolves around a physics problem involving a speeder and a cop car, where the speeder travels at a constant speed of 110 km/h and the cop car accelerates from rest at 9 km/h² until it reaches a maximum speed of 170 km/h. Participants are trying to determine how long it takes for the cop car to catch up with the speeder, with various attempts to relate their distances and times.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equal distances traveled by both vehicles and explore the necessary formulas to relate their speeds and times. There are questions about how to handle the conversion of units and the implications of acceleration on distance. Some participants suggest calculating the time it takes for the cop car to reach its maximum speed and the distance covered during that time.

Discussion Status

There is ongoing exploration of the problem with various participants offering insights and calculations. Some have provided partial calculations and suggestions for further steps, while others express confusion about unit conversions and the overall approach. Multiple interpretations of the problem are being considered, and participants are actively engaging with each other's ideas.

Contextual Notes

Participants are grappling with the conversion of speeds from km/h to km/s and the implications of acceleration on the distances traveled. There is also a mention of the initial conditions of the problem, including the cop car starting from rest and the need to determine the initial distance between the two vehicles.

  • #31
but what is that the time of? i am so lost right now
 
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  • #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 the 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 the person?

It is the same time. 17cos(50)*t gives the horizontal distance.
 
  • #51
ok so if i solve for t on that last equation u gave me...

.422= 17sin50 (t) = .5(-9.8)t squared

.422 = 13.022t + -4.9t squared

and then i get stuck in solving for t
 
  • #52
as u can tell my algebra is strugglin tonite
 
  • #53
anglum said:
ok so if i solve for t on that last equation u gave me...

.422= 17sin50 (t) = .5(-9.8)t squared

.422 = 13.022t + -4.9t squared

and then i get stuck in solving for t

use the quadratic equation.
 
  • #54
god i feel so dumb right now ... how do i use the quadratic equation to solve that

this is bad ... and I am so sorry
 
  • #55
anglum said:
as u can tell my algebra is strugglin tonite

no prob.
 
  • #56
i can't even function as to the quadratic equation solving that for me...
 
  • #57
anglum said:
god i feel so dumb right now ... how do i use the quadratic equation to solve that

this is bad ... and I am so sorry

that's ok. have a look at the quadratic formula here:

http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula

First arrange the equation in the correct form:

4.9t^2 -13.022t +0.422 = 0

now try to apply the quadratic formula using the formula in the link.
 
  • #58
t = 2.62?
 
  • #59
anglum said:
t = 2.62?

yes. looks right.
 
  • #60
so then for the horizontal distance it is just 17 cos 50 (2.62)?
 

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