Recent content by dizweld

Thanks to the both of you for the help. I will find out what he expected once it is graded I guess ^ ^.

I tried this once rearranging equation to -Fdrag = ma - Fengine= 1264.5N-8487N= -7222.5N It seemed too much to me, so maybe I am not rearranging properly. Cant see another way to do it, can it be really that simple an answer?

I have with no luck unfortunately, not sure how to use that.

My prof. has stated it's possible to use kinematics/motion for this? If that helps at all. Yeah this problem is very frustrating to me.

"A 1500kg car accelerates in first gear to from 0 to 55km/hour in 2.7 seconds. In 5th gear the car accelerates from 150km/h to 220km/h in 23 seconds. Assume the drag force from wind is negligible at low speeds but increases as the square of the velocity. Assuming there is negligible wind drag in...

So alright here is what I'm doing. 8487N*(15.277m/s)=129662.5P If the power is the same at a velocity of 220km/h->61.11m/s Then I can do F=P/velocity, but that gives me =2121.788N? Which is higher than 1267.5N the sum of engine and drag. Is that to say engine power is 2121N and drag is the...

Tried employing work energy theorem but I'm not there yet so...trying to figure out how to do it another way, difficult for me heh.

8487n.

I see that now wow. I was, still am confused by "increases by square of velocity".

Assuming 0-55km/h is low speed and 150 to 220km/h is high speed, so I calculated the acceleration from 150 to 220 multiplied it by 1500kg after conversion and got that force. edit: 0-55 in 2.7 seconds is a quick acceleration hinting that there is negligible drag force...The only data left is...

What about this? So if square of velocity is the force of drag, then (200km/h)^2--->(55.55m/s)^2= 3086.4N Then since apparently engine force is 1267.5N, 1267.5N-3086.4N= -1818.9N I know I'm probably wrong...If I divided by mass that would give acceleration but I don't think that is drag...

Oh sorry. Well a=f/m , and both force due to engine and drag are forces so f=a*m

Has to be mass right? Maybe velocity is a better answer.

Well force of drag goes opposite direction of force of car, so in order to accelerate to higher velocities more force car is required to achieve that acceleration? Hm

My issue/knowledge Hello all, I think this is my first time posting a question here (I have read the guidelines). This problem is not in my textbook or notes, and I'm not sure how to go about it. Online all I can find about air drag equations requires knowing variables which I'm not given here...