
#1
May1506, 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 



#2
May1506, 09:58 PM

PF Gold
P: 7,125

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
May1506, 10:02 PM

Sci Advisor
HW Helper
P: 6,590

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



#4
May1506, 10:08 PM

P: 34

Newton Law: Force
thanks for the help guys really appreciate it




#5
May1506, 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' 



#6
May1506, 11:18 PM

Sci Advisor
HW Helper
P: 6,590

AM 



#7
May1506, 11:23 PM

HW Helper
P: 2,887

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
May1506, 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




#9
May1506, 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?




#10
May1606, 01:20 AM

Sci Advisor
HW Helper
P: 6,590

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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