Homework Help: Could use some help on these problems! (average velocity)

1. Sep 9, 2006

wtf_albino

Hey everyone! i'm new to the forums

I just started AP Physics a few days ago and i'm having some trouble on the following problems:

1) A truck on a straight road starts from rest accelerating at 2.0 m/s^2 until it reaches a speed of 20 m/s. Then the truck travels for 20 s at a constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 s. (A)How long is the truck in motion? (B)What is the average velocity of the truck for the motion described?
__________________________________________________ ____________

For (A) I found the correct time to be 35 s.

For (B), I used the equation:

Average velocity = (Vo + V)/2

I set it up like:

V = [(0 m/s) + (20 m/s)] / 2

i came up with 10 m/s as my velocity which was wrong. I know this is kind of a multifaceted problem and i'm having trouble adjusting the equation accordingly. Can someone please point out my error and how i can solve this?

__________________________________________________ ____________
2) A cesna aircraft has a lift-off speed of 120 km/h. (A) What minimum constant acceleration does this require if the aircraft is to be airborne after a takeoff run of 240 m? (B) How long does it take the aircraft to become airborne?
_____________

For this problem i used the equation:

V^2 = Vo^2 + 2ax

I set it up like:

(120 km/h)^2 = (0 km/h)^2 + 2(a)(.240 km)

i came up with 30,000 km/h^2 which doesn't sound right at all

Can someone please point out my error and how i am able to better approach this problem?

Help would be very much appreciated thanks!

2. Sep 9, 2006

HallsofIvy

At 2 m/s2, its speed at time t is 2t m/s= 20 m/s so t= 10 s- that is, it take 10 seconds to reach a speed of 20 m/s. You are told that the truck travels at that speed for 20 s and then decelerates to a stop in 5 s. Yes, the truck is in motion for 10+ 20+ 5= 35 s.

That's the average velocity while the vehicle was accelerating. It does not include the time at constant speed or while decelerating.

Then do the different "facets". "Average velocity" is total distance traveled divided by the time. You have already determined that the time of motion was 35 s so you need to find the total distance traveled.

You can calculate the total distance traveled while accelerating in two different ways. Since the distance traveled, starting from 0 speed at constant acceleration a, is (1/2)at2, at 2 m/s2 for 10 seconds the car goes (1/2)(2)(10)2= 100 m. Or, as you calculated, since the car starts at 0 and accelerates up to 20 m/s, its average velocity was (0+ 20)/2= 10 m/s. At that average speed for 10 s, it goes 100 m.
Then the car went at 20 m/s for 20 s. It traveled 20(20)= 400 m in that time.
Finally, the car went from 20 m/s to 0, an average velocity of 10 m/s, for 5 seconds- it went 10(5)= 50 m while decelerating. That means the car went 100+ 400+ 50= 550 m in 35 s, an average velocity of 550/35= 15.7 m/s, approximately.

I'm not certain about that formula. It may well be correct but I'm not familiar with it. Here's how I would do the problem: at constant acceleration a, after time t, the speed would be v= at and the distance traveled would be (1/2)at2. Here you have v= at= 120 km/h and x= (1/2)at2= .240 km. From at= 120, t= 120/a. Putting that into the second equation, (1/2)at2= (1/2)a(14400/a2= 7200/a= 0.24. Then a= 7200/.24= 30,000 km/h2. Yes, that is exactly what you got. If "two brilliant minds" get the same answer .... Of course, Knowing the intitial velocity= 0 and final velocity= 120, the average velocity is (0+120)/2= 60 km/h so the time required to go that 0.24 km is 0.24/60= 0.004 h= 0.24 m= 14.4 s.

3. Sep 9, 2006

wtf_albino

thanks a lot for clearing that up for me halls, i really appreciate it :D