Solving Kinematics Problems: Distance Traveled Before Applying Brakes

In summary, the average alcohol-free driver requires about 0.8 s to apply the brakes after seeing an emergency.
  • #1
nblu
56
0
Q: The average alcohol-free driver requires about 0.8 s to apply the brakes
after seeing an emergency. Calculate the distance traveled after seeing the emergency
before applying the brakes.

with the question, the given speed is 17 m/s and t = 0.8.
at first, i used d =vt equation to find the distance, however,
when i read through the question again, i had a feeling that it's not right.
my first answer was d=14m, but it just doesn't look correct.

i was thinking of finding the acceleration then use v[tex]^{2}_{f}[/tex]=v[tex]^{2}_{i}[/tex] +2a[tex]\Delta[/tex]d
to solve for d, OR, use this equation, [tex]\Delta[/tex]d=1/2(v[tex]_{1}[/tex]+v[tex]_{2}[/tex])[tex]\Delta[/tex]t, without
even having to find the acceleration.

one part of my knowledge tells me that i don't need acceleration to do this question
because this is about "before" applying the break.

its confusing me..

any help would be greatly appreciated.

thank you
 
Last edited:
Physics news on Phys.org
  • #2
You "know" what you did is right. Resist the temptation to complicate it. I'm sure you'll get a chance to use that other stuff later.
 
  • #3
nblu said:
with the question, the given speed is 17 m/s and t = 0.8.
at first, i used d =vt equation to find the distance, however,
when i read through the question again, i had a feeling that it's not right.
my first answer was d=14m, but it just doesn't look correct.


one part of my knowledge tells me that i don't need acceleration to do this question
because this is about "before" applying the break.

If Dick's reply isn't enough, your logic is right to me, simple enough. I don't see how it doesn't "look correct".
 
  • #4
Dick said:
You "know" what you did is right. Resist the temptation to complicate it. I'm sure you'll get a chance to use that other stuff later.

Siracuse said:
If Dick's reply isn't enough, your logic is right to me, simple enough. I don't see how it doesn't "look correct".

Thank you, I really appreciate it! :)
 

FAQ: Solving Kinematics Problems: Distance Traveled Before Applying Brakes

What is kinematics and how is it related to physics?

Kinematics is a branch of physics that studies the motion of objects without considering the forces that cause the motion. It deals with concepts such as position, velocity, acceleration, and time.

How do I solve kinematics problems?

To solve a kinematics problem, you need to first identify the known and unknown quantities, and then use the appropriate equations of motion to calculate the unknown quantity. It is important to draw a diagram and label all the given information to help visualize the problem.

What are the three equations of motion in kinematics?

The three equations of motion in kinematics are:

1. v = u + at - relates final velocity (v), initial velocity (u), acceleration (a), and time (t).

2. s = ut + 1/2at2 - relates displacement (s), initial velocity (u), acceleration (a), and time (t).

3. v2 = u2 + 2as - relates final velocity (v), initial velocity (u), acceleration (a), and displacement (s).

What is the difference between average velocity and instantaneous velocity?

Average velocity is the average rate of change of an object's position over a given time interval, while instantaneous velocity is the velocity of an object at a specific moment in time. Average velocity can be calculated by dividing the total displacement by the total time, while instantaneous velocity can be calculated by taking the derivative of the position-time graph at a specific time.

Can kinematics be applied to both linear and rotational motion?

Yes, kinematics can be applied to both linear and rotational motion. In linear motion, objects move in a straight line, while in rotational motion, objects rotate about an axis. The same concepts of displacement, velocity, and acceleration can be applied to both types of motion, but the equations of motion may differ slightly.

Similar threads

Back
Top