# Homework Help: A car travelling at 50km/h stops in 70.0m HELP

1. Mar 8, 2014

### lozah

1. The problem statement, all variables and given/known data

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

2. Relevant 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.

3. 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...?

2. Mar 8, 2014

### lozah

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?

3. Mar 8, 2014

### Simon Bridge

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.

4. Mar 9, 2014

### CWatters

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.

5. Mar 9, 2014

### SammyS

Staff Emeritus