Kinematic Equation Problem using Velocity & Distance

In summary, the problem is to find the no-skid stopping distance of a car with an initial speed of 26.9 m/s and an acceleration of -8.2 m/s^2, assuming a reaction time of 0.50 s. The correct answer is 57.57 m, but the incorrect answer of 44.12 m was obtained by only using the formula for motion under constant acceleration and not accounting for the distance traveled during the reaction time.
  • #1
dlang
1
0
Hey guys, I'm confused on this problem: It is a typical sunny day. Assume your reaction
time is 0.50 s and the acceleration of your car is –8.2 m/s2. What is the no-skid stopping distance when the car’s speed is initially 26.9 m/s?

Homework Statement


a=-8.2 m/s^2
v0=26.9 m/s
v=0
t=0.50s
d=?


Homework Equations



d=v0(t) +1/2a(t^2)
v^2=v0^2+2ad


The Attempt at a Solution


So I tried using the v^2=v0^2+2ad and got 44.12 m. However, the answer is 57.57. What did I do wrong? Thanks!
 
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  • #2
dlang said:
Hey guys, I'm confused on this problem: It is a typical sunny day. Assume your reaction
time is 0.50 s and the acceleration of your car is –8.2 m/s2. What is the no-skid stopping distance when the car’s speed is initially 26.9 m/s?

Homework Statement


a=-8.2 m/s^2
v0=26.9 m/s
v=0
t=0.50s
d=?

Homework Equations



d=v0(t) +1/2a(t^2)
v^2=v0^2+2ad

The Attempt at a Solution


So I tried using the v^2=v0^2+2ad and got 44.12 m. However, the answer is 57.57. What did I do wrong? Thanks!
Hello dlang. Welcome to PF !

Show detail regarding how you used the equations you gave.
d = v0(t) +(1/2)a(t2)
v2 = v02+2ad​
 
  • #3
dlang said:
Hey guys, I'm confused on this problem: It is a typical sunny day. Assume your reaction
time is 0.50 s and the acceleration of your car is –8.2 m/s2. What is the no-skid stopping distance when the car’s speed is initially 26.9 m/s?

Homework Statement


a=-8.2 m/s^2
v0=26.9 m/s
v=0
t=0.50s
d=?


Homework Equations



d=v0(t) +1/2a(t^2)
v^2=v0^2+2ad


The Attempt at a Solution


So I tried using the v^2=v0^2+2ad and got 44.12 m. However, the answer is 57.57. What did I do wrong? Thanks!

What you have used is the formula for motion under constant acceleration which is true only after you have reacted. What about the distance traveled during your reaction time?
 

1. What are the kinematic equations used to solve problems involving velocity and distance?

The three most commonly used kinematic equations to solve problems involving velocity and distance are the displacement equation (s = v0t + 1/2at2), the final velocity equation (v = v0 + at), and the average velocity equation (v = (v0 + vf)/2).

2. How does one determine which kinematic equation to use in a problem?

The kinematic equation used in a problem depends on the given information and what is being solved for. If the initial and final velocities are known, the final velocity equation can be used. If the displacement, initial velocity, and time are known, the displacement equation can be used. If the initial and final velocities and the time are known, the average velocity equation can be used.

3. Can kinematic equations be used to solve problems with constant acceleration?

Yes, the kinematic equations are specifically designed to solve problems with constant acceleration. In fact, the equations assume that the acceleration remains constant throughout the motion.

4. What units should be used when plugging in values to the kinematic equations?

The units used in each kinematic equation should be consistent. For example, if the displacement is given in meters and the time is given in seconds, the initial and final velocities should also be in meters per second, and the acceleration should be in meters per second squared.

5. Can the kinematic equations be used to solve problems involving non-uniform motion?

No, the kinematic equations are only applicable to problems with constant acceleration. If the acceleration is not constant, a different approach, such as using calculus, is needed to solve the problem.

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