Calculating Force: Newton's Law and the Motion of a 850 kg Dragster

In summary, the conversation discusses two questions about a dragster starting from rest and attaining a speed of 24.6m/s in 0.49s. The first question asks for the average force on the dragster during this time interval, which is found using the equation F=ma. The second question asks for the horizontal force exerted on the driver, which is found by using the driver's mass and acceleration. The conversation then moves on to another question about finding the acceleration and final velocity of the dragster, which can be solved using the equations of motion under constant acceleration.
  • #1
aerogurl2
34
0
I need some help on the two questions:
A 850 kg dragster, starting from rest, attains a speed of 24.6m/s in 0.49s.

1. What is the size of the average force on the dragster during this time interval?

Do I just use the equation F=ma to find this answer?...because I found the acceleration which is 50.2...so I just plug in the numbers F=(850)(50.2) which equals 42670? :confused:

2.Assume that the driver has a mass of 83kg. What horizontal force does the seat exert on the driver? I don't get this what do they mean by horizontal force and how do you find it??

Please someone help me:cry: thanx
 
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  • #2
Yup, you did part 1 correctly

Part 2 is basically asking you what force was exherted on the driver (horizontal is specified because you're not looking for the vertical normal force). Do it just like the first one
 
  • #3
aerogurl2 said:
I need some help on the two questions:
A 850 kg dragster, starting from rest, attains a speed of 24.6m/s in 0.49s.

1. What is the size of the average force on the dragster during this time interval?

Do I just use the equation F=ma to find this answer?...because I found the acceleration which is 50.2...so I just plug in the numbers F=(850)(50.2) which equals 42670?
What are the units of force?

[tex]f_{avg} = ma_{avg} = m\Delta v/\Delta t[/tex]

2.Assume that the driver has a mass of 83kg. What horizontal force does the seat exert on the driver? I don't get this what do they mean by horizontal force and how do you find it??
What is the acceleration of the driver? What is his mass? What is the force then?

AM
 
  • #4
thanks for the help guys really appreciate it:smile:
 
  • #5
I have one more question.. the dragster completed the 412.3m run in 4.936s. if the car had a constant acceleration, what would be the acceleration and final velocity?
I only know how to find the average velocity by dividing 412.3m by 4.936s but I don't know how to find the final velocity and the acceleration b/c they're not asking for the average velocity or average acceleration

I thought I have to plug the numbers into an equation like this one: vf=a(t) but I don't know either 'a' or 'Vf'
 
  • #6
aerogurl2 said:
I have one more question.. the dragster completed the 412.3m run in 4.936s. if the car had a constant acceleration, what would be the acceleration and final velocity?
I only know how to find the average velocity by dividing 412.3m by 4.936s but I don't know how to find the final velocity and the acceleration b/c they're not asking for the average velocity or average acceleration

I thought I have to plug the numbers into an equation like this one: vf=a(t) but I don't know either 'a' or 'Vf'
Draw a graph of velocity vs. time where acceleration is constant. What does it look like? What is the area under the graph in terms of a and final t? What does the area represent? So how is the distance related to the a and t? If you can determine that, you can find t. When you know t, you can find final v (what is relationship between v, a and t?).

AM
 
  • #7
aerogurl2 said:
A 850kg dragster, starting from rest, attains a speed of 24.6 m/s in 0.49s.

the dragster completed the 412.3m run in 4.936s. if the car had a constant acceleration, what would be the acceleration and final velocity?
I only know how to find the average velocity by dividing 412.3m by 4.936s but I don't know how to find the final velocity and the acceleration b/c they're not asking for the average velocity or average acceleration

to get the acceleration do i do f/m=a? or is it just v/t?
can someone try helping me?

Are you familiar with the equations of motion under constant acceleration?

You need these two for this problem :

[tex]v = u + at[/tex] where v is final velocity, u is initial velocity (zero in this case since it starts from rest), a is acceleration and t is time. Initially, you're given the speed the dragster reaches after a certain amount of time under constant acceleration. Use this to find out a.

[tex]v^2 = u^2 + 2as[/tex] where s is displacement (distance from the start). Plug in the previously calculated value of a to find v after the full distance is covered.
 
  • #8
after I graph the velocity vs. time graph I understand that the line has to be linear for the acceleration to be constant the area under the curve in the velocity graph is 1017.56 this shows the total distance travel but I don't understand how knowing the total distance travel can help me find constant acceleration Please advise me on this question
 
  • #9
after looking at it the equation for the velocity graph is y= 412.3/4.936x therefore in the acceleration graph it has to show a horizontal line which is the slope of the velocity graph therefore the constant acceleration is 83.53m/s^2. Am I on the right track?
 
  • #10
aerogurl2 said:
after looking at it the equation for the velocity graph is y= 412.3/4.936x therefore in the acceleration graph it has to show a horizontal line which is the slope of the velocity graph therefore the constant acceleration is 83.53m/s^2. Am I on the right track?
The essential point is that the area under the graph (distance travelled) is 1/2 the final velocity x elapsed time.

The area under the graph at time t is the area of a triangle of base t and height of v. So the area (which is the distance travelled) is:

(1)[tex]Area = s = \frac{1}{2}vt[/tex]

But where a is constant: [itex]v = at[/itex] so

(2)[tex]s = \frac{1}{2}at^2[/tex]

So work out a from (2) (you know t and s). Then work out final v from (1).

AM
 

1. What is Newton's second law of motion?

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be expressed as F=ma, where F is the net force, m is the mass of the object, and a is the acceleration.

2. How do you calculate the force required to accelerate an 850 kg dragster?

To calculate the force required to accelerate an 850 kg dragster, we can use the formula F=ma, where F is the force, m is the mass of the dragster (850 kg), and a is the desired acceleration. For example, if we want to accelerate the dragster at 10 meters per second squared, the force required would be 8,500 newtons (N).

3. What is the relationship between force and acceleration?

The relationship between force and acceleration is directly proportional. This means that as the force acting on an object increases, its acceleration also increases. Similarly, if the force decreases, the acceleration decreases as well.

4. Can you explain how Newton's third law of motion applies to the 850 kg dragster?

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In the case of the 850 kg dragster, as the dragster moves forward and exerts a force on the ground with its wheels, the ground also exerts an equal and opposite force on the dragster, propelling it forward.

5. How does friction affect the force required to accelerate the dragster?

Friction is a force that opposes motion, and it affects the force required to accelerate the dragster. The dragster's tires experience friction with the ground, which can reduce the net force acting on the dragster and thus affect its acceleration. This is why dragsters have specially designed tires with less friction for better acceleration.

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