Basic problem on the equation of motion

In summary, the conversation is discussing how to calculate the braking force and time for a car to come to rest, given its mass, initial velocity, and distance traveled. Through using kinematics equations and correcting for units, the correct answers are determined to be 1.87kN for the braking force and 6.55 seconds for the time. The formula for calculating acceleration using distance traveled is also discussed and used to verify the results.
  • #1
Frisky90
19
0

Homework Statement



The brakes on a car are applied when traveling at 55km/h and it comes to rest uniformly in a distance of 50m. The mass of the car is 800 kg . Calculate the braking force and the time for the car to come to rest.

Homework Equations



F=ma
v=u+at
t=s/v=50*3.6/55=3.27s

0=55/3.6 +a*3.27
a=-4.67 deceleration. Then F=800*4.67=3738.2263N braking force
However, according to the txtbook the right answers are: 1.87kN, 6.55s ?
 
Physics news on Phys.org
  • #2
Frisky90 said:

Homework Statement



The brakes on a car are applied when traveling at 55km/h and it comes to rest uniformly in a distance of 50m. The mass of the car is 800 kg . Calculate the braking force and the time for the car to come to rest.

Homework Equations



F=ma
v=u+at
t=s/v=50*3.6/55=3.27s

0=55/3.6 +a*3.27
a=-4.67 deceleration. Then F=800*4.67=3738.2263N braking force
However, according to the txtbook the right answers are: 1.87kN, 6.55s ?

Watch your units...your kinematics equations deal with m/s and m/s/s. You have km/h.

Also, from an energy standpoint, I verified that 1.87kN is correct for the force.
 
  • #3
Apphysicist said:
Watch your units...your kinematics equations deal with m/s and m/s/s. You have km/h.

Also, from an energy standpoint, I verified that 1.87kN is correct for the force.

I can't find a problem with the units.I've converted km/h to m/s.
And I know that the textbook is correct :)
 
  • #4
You have used distance traveled inappropriately. The correct way to plug in distance is
a = (v^2 - u^2) / 2s
where v is final velocity, u is initial velocity, and s is distance traveled.
 
Last edited:
  • #5
Chronos said:
You have used distance traveled inappropriately. The correct way to plug in distance is
a = (v^2 - u^2) / 2s
where v is final velocity, u is intial velocity, and s is distance traveled.

I don't understand how do you derive this expression for a? Could you explain? And if you plug in the numbers in it, you still don't get the correct answer.
 
  • #6
Permit me to clarify. Distance traveled is the area under the line of the time - velocity graph. Since this is a straight line when acceleration is constant, the area under the line is a simple triangle, so the formula is
s = [(v + u)*t] / 2 or
t = 2s / (v + u)

for s =50, v = 0 and u = 15.273

t = 100/15.273 = 6.548 seconds.

solving for [a] we use v = u + at and get

0 = 15.273 + at
0 = 15.273 + 6.548a
6.548a = -15.273
a = -2.332

The results are the same using the expression I originally gave, albeit that one is a little harder to derive.
 

What is the equation of motion?

The equation of motion is a mathematical expression that describes the relationship between an object's position, velocity, and acceleration over time. It is commonly written as S = ut + (1/2)at^2, where S is the object's displacement, u is its initial velocity, a is its acceleration, and t is time.

What is the difference between displacement and distance?

Displacement is the straight-line distance between an object's initial and final positions, while distance is the total path length traveled by the object. For example, if an object moves 10 meters east and then 5 meters west, its displacement is 5 meters east, but its distance traveled is 15 meters.

What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both speed and direction. In other words, velocity takes into account the object's movement in a specific direction, while speed does not. For example, a car traveling at a constant speed of 60 mph north has a velocity of 60 mph north, while a car traveling at a constant speed of 60 mph in a circular path has a velocity of 60 mph but is constantly changing direction.

How is acceleration related to the equation of motion?

Acceleration is the rate of change of an object's velocity over time. In the equation of motion, acceleration is represented by the term at^2, which shows that the object's acceleration is directly proportional to the square of the time. This means that the longer an object accelerates, the greater its change in velocity will be.

How can the equation of motion be applied in real-life situations?

The equation of motion can be used to predict an object's future position and velocity based on its starting conditions. This is particularly useful in physics and engineering, where it can be used to design and analyze the motion of objects such as cars, airplanes, and projectiles. It can also be used in everyday situations, such as calculating the time it will take to reach a destination based on your speed and distance traveled.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
630
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
823
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
734
Back
Top