A car travelling at 50km/h stops in 70.0m HELP

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Homework Help Overview

The problem involves a car that stops from a speed of 50 km/h over a distance of 70 m, with a question about determining the stopping distance at a higher speed of 90 km/h. The context is rooted in the concepts of momentum and energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using ratios to relate stopping distances at different speeds, with one participant attempting to apply a squared relationship between speed and stopping distance. Others suggest using graphical representations and energy equations to analyze the problem.

Discussion Status

There are multiple lines of reasoning being explored, including the use of ratios, graphical analysis, and energy equations. Some participants express confusion about the application of formulas, while others offer insights into potential mathematical approaches without reaching a consensus.

Contextual Notes

Participants are navigating the complexities of relating speed to stopping distance, with some questioning the applicability of certain formulas and concepts from their studies on momentum and energy.

lozah
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Homework Statement



A car traveling at 50.0 km/h stops in 70.0 m. What is the stopping distance if the car's speed is 90 km/h?


Homework Equations



This is a question in the unit about momentum & energy, so most of the questions have been applying formulae about momentum, work, gravitational potential energy, kinetic energy, etc. I'm confused as to how this question relates to this section because I can't seem to find an applicable formula.

The Attempt at a Solution



I attempted to solve this using ratios and got an answer of 126 m, but the real answer is somewhere around 2.3E2 m. I'm super confused, but expecting that it'll be a super obvious solution...?
 
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OH! Maybe I figured it out. Since doubling velocity QUADRUPLES the stopping distance (because 2^2 = 4), and the velocity was increased by 1.8, I can square 1.8 (1.8^2 = 3.24) and multiply 3.24 by 70.0 to get 226.8! Does anyone know how I could show this in a more mathematical way?
 
You can also see it in terms of the v-t graph.
Model the car as stopping with a constant acceleration ... this is a line on your graph with a negative slope - so it makes a triangle with the v and t axes. The area inside this triangle is the distance to stop.

Sketch the graph for both situations, then you have two similar traingles - don't let the time to stop T being unknown phase you - what's important is that the acceleration is the same each time.

If you prefer to memorize equations - the equations you want are the kinematic or "suvat" equations.
You want the one that relates displacement to velocity and acceleration.
 
I was going to suggest writing two energy equations using..

Work = Force * distance = 0.5mV2

eg

F * D1 = 0.5mV12

F * D2 = 0.5mV22

Substitute for F.
The mass will cancel.
 
lozah said:
OH! Maybe I figured it out. Since doubling velocity QUADRUPLES the stopping distance (because 2^2 = 4), and the velocity was increased by 1.8, I can square 1.8 (1.8^2 = 3.24) and multiply 3.24 by 70.0 to get 226.8! Does anyone know how I could show this in a more mathematical way?
I like your approach here.
 

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