Am I setting up this kinematic equation properly?

Click For Summary
SUMMARY

The discussion focuses on calculating the acceleration, final speed, and stopping distance of a race car under constant acceleration. The car accelerates at 5.5 m/s², reaching a final speed of 66.5 m/s after traveling a quarter mile (approximately 402.3 m) in 12.1 seconds. For the stopping distance, the user initially calculated 1507.8 m using the wrong initial position, later correcting it to 1105.5 m after resetting the initial position to zero. The braking acceleration of -2.00 m/s² is identified as modest, with safe braking typically around 0.5g.

PREREQUISITES
  • Understanding of kinematic equations
  • Knowledge of acceleration and deceleration concepts
  • Familiarity with unit conversions (miles to meters)
  • Basic physics principles related to motion
NEXT STEPS
  • Study the kinematic equation: v² = v₀² + 2a(x - x₀)
  • Learn about the effects of different braking accelerations on stopping distances
  • Research safe braking distances for various vehicles under different conditions
  • Explore the concept of g-forces in relation to acceleration and braking
USEFUL FOR

Physics students, automotive engineers, race car enthusiasts, and anyone interested in vehicle dynamics and motion analysis.

TaylorHoward21
Messages
10
Reaction score
1
A race car starts from rest and goes a quarter mile (1/4 mi) in 12.1 seconds. Assume the acceleration of the car is constant. (a.) What is the acceleration of the car? (b.) What is the final speed of the car? (c.) If you hit the brakes, how far would it take to stop the car if the breaks cause an acceleration of -2.00 m/s2?

I believe I have parts a and b right, however I am not confident with my process of solving part C.


1/4 mi ≈ 402.3 m

a.) acceleration = 5.5m/s^2
b.) Final speed (at 12.1 seconds) = 66.5 m/s

c.) For part C, I reset the known values to:
a = -2.00m/s2
V0 = 66.5m/s
V = 0 (because solving for car stop which is zero velocity)
X0 = 402.3m

The equation I used: v2 - V20 = 2a(x - x0)
The answer I got was 1507.8m = x
For a car going 66.5m/s that is slowing down at -2.00 m/s2 does this answer seem practical? I am struggling to check my work here.

Many thanks.
 
Physics news on Phys.org
TaylorHoward21 said:
X0 = 402.3m
It asks how far the car goes after hitting the brakes. The distance already traveled is not relevant.
 
haruspex said:
It asks how far the car goes after hitting the brakes. The distance already traveled is not relevant.
So if I change X0 to be = 0 the new answer I get is 1105.5m. This still seems rather far for a car to stop from 66.5m/s to 0m/s. Is -2.00m/s^2 simply just a very slow breaking acceleration?
 
TaylorHoward21 said:
Is -2.00m/s^2 simply just a very slow breaking acceleration?
It is rather modest. Safe but serious braking is about 0.5g on a dry road. Emergency braking could go as high as 0.7-1g, depending on the vehicle.

And it is "braking", using "brakes", not "breaking".
 
  • Like
Likes   Reactions: Orodruin

Similar threads

  • · Replies 34 ·
2
Replies
34
Views
3K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
2
Views
4K
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
11K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
7K