Kinematics - Constant Deceleration

In summary, the problem involves finding the reaction time and deceleration of a car based on the total distance traveled during two phases of braking. The equations used to solve this problem are based on constant acceleration and require setting all units equal to each other.
  • #1
rtsphysics
5
0

Homework Statement


To stop a car, you require first a certain reaction time to begin braking; then the car slows down under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 186 ft when its initial speed is 50 mi/hr, and 80 ft when its initial speed is 30 mi/hr. What are (a) your reaction time and (b) the magnitude of the deceleration?

Problem taken from Fundamentals of Physics 5th ed. Halliday, Resnick, Walker Pg. 32, prob. 51P


Homework Equations



x - x0 = 0.5(v0 + v)t
x - x0 = vt - 0.5at2
v2 = v02 + 2a(x - x0)


The Attempt at a Solution


Ok so I tried plugging in the displacements 186 ft and 80 ft into the above equations. The acceleration is constant so I can use the equations above. However, when i plug it in, I always get two or more equations with at least two variables. For example, plugging both numbers in the first equation, we get:

186 ft = 0.5(50mi/hr + v1)t1
80 ft = 0.5(30mi/hr + v2)t2

I don't see how I can get one of the unknown variables such as final velocity using any other equations, such as v = v0 + at, since i don't know a or t...

Any help would be appreciated!
 
Physics news on Phys.org
  • #2
you don't need an equation for the final velocity ,for it is given by statement of "to stop a car".
 
  • #3
Try setting all your units equal to each other. Make sure you are not setting feet equal with miles and hours with seconds. Convert feet to miles or miles/hr to feet/hr. All units of time should be the same unit and all the units of distance should be the same.
 

Related to Kinematics - Constant Deceleration

1. What is constant deceleration in kinematics?

Constant deceleration in kinematics refers to a situation where an object's velocity decreases at a constant rate over time. This means that the object is slowing down at the same amount in each unit of time.

2. How is constant deceleration different from constant acceleration?

Constant deceleration is the opposite of constant acceleration. In constant acceleration, an object's velocity increases at a constant rate, whereas in constant deceleration, an object's velocity decreases at a constant rate.

3. What is the formula for calculating displacement with constant deceleration?

The formula for calculating displacement with constant deceleration is d = v0t - 1/2at2, where d is displacement, v0 is initial velocity, t is time, and a is the deceleration rate.

4. Can an object have both constant acceleration and constant deceleration?

No, an object cannot have both constant acceleration and constant deceleration at the same time. This is because acceleration and deceleration are opposite processes, and only one can occur at a time. However, an object can have varying acceleration or deceleration rates at different points in its motion.

5. How does air resistance affect an object's constant deceleration?

Air resistance can affect an object's constant deceleration by slowing down the object's velocity. As the object moves through the air, the air resistance creates a force in the opposite direction of motion, causing the object to decelerate at a faster rate than it would without air resistance.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
33
Views
4K
  • Introductory Physics Homework Help
Replies
29
Views
7K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
34
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
29
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
5K
Back
Top