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Newton Law: Force

by aerogurl2
Tags: force, newton
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aerogurl2
#1
May15-06, 09:53 PM
P: 34
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???

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 thanx
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Pengwuino
#2
May15-06, 09:58 PM
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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
Andrew Mason
#3
May15-06, 10:02 PM
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Quote Quote by aerogurl2
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

aerogurl2
#4
May15-06, 10:08 PM
P: 34
Newton Law: Force

thanks for the help guys really appreciate it
aerogurl2
#5
May15-06, 10:31 PM
P: 34
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'
Andrew Mason
#6
May15-06, 11:18 PM
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Quote Quote by aerogurl2
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
Curious3141
#7
May15-06, 11:23 PM
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Quote Quote by aerogurl2
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.
aerogurl2
#8
May15-06, 11:36 PM
P: 34
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
aerogurl2
#9
May15-06, 11:46 PM
P: 34
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?
Andrew Mason
#10
May16-06, 01:20 AM
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Quote Quote by aerogurl2
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


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