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

  • Thread starter Thread starter lozah
  • Start date Start date
  • Tags Tags
    Car
Click For Summary
SUMMARY

A car traveling at 50.0 km/h stops in 70.0 m, and when the speed increases to 90 km/h, the stopping distance can be calculated using the principle that doubling the velocity quadruples the stopping distance. The correct stopping distance for 90 km/h is approximately 226.8 m, derived by squaring the velocity ratio (1.8) and multiplying by the original stopping distance. The kinematic equations, particularly the one relating displacement to velocity and acceleration, are essential for this calculation.

PREREQUISITES
  • Understanding of kinematic equations (SUVAT)
  • Knowledge of momentum and energy concepts
  • Familiarity with graphing velocity-time relationships
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the kinematic equations for displacement, velocity, and acceleration
  • Learn about the relationship between kinetic energy and stopping distance
  • Explore the concept of constant acceleration in physics
  • Practice solving problems involving momentum and energy
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of kinematic principles in action.

lozah
Messages
5
Reaction score
0

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...?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
2K
Replies
7
Views
2K
Replies
9
Views
16K