PLEASE HELP WITH THIS PHYSICS PROBLEM!! test tomorrow and super cofused :( :( neeed help a.s.ap?

Question

hey
i have a test tomorrow on problems about acceleration and velociy and stuff and i have no clue how to do this stuff :( can someone please help me and explain?

a plan starting from rest at one end of a runway, undergoes a constant accerleation of 1.6 m/s2 for a distance of 1600m before takeoff. what is its speed upon takeoff? what is the time required for takeoff?

i tried plugging this in..
do i use the acceleration formula?
but switch it around so the time variable is alone?
omgosh please help and explain

Answer 1

Don`t panic
1.) Start with writing down the given parameter:

acceleration a=1.6 m/s^2
distance s=1600 m
initial velocity v0=0
initial distance s0=0

2.) write down the unnown parameter:

Final velocity v1
time at final velocity t

3.) decide what kind of motion you have (accelerated or not accelerated) and
4.) choose the formula depending on 3.)

In your example:

s=1/2 a*t^2
t=sqrt[2*1600 m/(1.6 m/s^2)]= 44.72 s

v=a*t=(1.6 m/s^2)*44.72 s = 71.55 m/s

good luck for tomorrow!

Answer 2

v^2 - u^2 = 2as is the formula to be used where u is the initial velocity, v the final velocity, a the acceleration and s the distance traveled.

v^2 - 0^2 = 2 x 1.6 x 1600 = 5120

v = sqrt.5120 = 71.55 m/sec.

Basically there are three formulas:

v = u + at

s = ut + 1/2 at^2

v^2 - u^2 = 2as

Answer 3

you need to actually use the motions formula
v2=u2+2*a*s [ S is the distance traveled ; v the final velocity and u the initial velocity]
v2=0+2*1.6*1600[ since the plane is initially at rest]
v2=5120
v= sqr. root of 5120
v=71
the answer is quite absurd but anyway thats the answer

Answer 4

use v^2 -u^2 =2as
where u=initial speed=0
v=final speed at take off
a=accelaration=1.6m\s2
s=length of the runway
to find time, use v=u+at with the value of v from above

Answer 5

simplify: (ax+by-bx-ay)/(ax-by+bx-ay) let's look at each part separately. ax+by-bx-ay this can be rewritten as ax-ay+by-bx and therefore as a(x-y) - b(x-y), and so (a-b)(x-y) ax-by+bx-ay can be rewritten as ax-ay+bx-by and therefore as a(x-y)+b(x-y), so (a+b)(x-y) so put it back together and you get [(a-b)(x-y)]/[(a+b)(x-y)] The (x-y) in the numerator and denominator cancel out and you are left with (a-b)/(a+b) ok, so if you have X^2(X^2-9)>0 so I know X =0 and X= + or - 3 but when graphing in terms of double roots. Is 3 and or 0 a double root? thankss 0 is a double root, but 3 is not. That is because when you have x^2 = 0, you have +/- 0, or 0 twice, but when you have x^2 = 9, you have +/- 3, or two separate solutions

Answer 6

(V0) initial velocity = 0
(V1) final velocity = X
acceleration = 1.6
distance = 1600

2ad + V0^2 = V1^2

(2*1.6*1600) + 0 = V1^2
sqrt (2*1.6*1600) = V1

(m/s^2)*(m) = (m/s)^2
meter per seconds squared times meters = meter squared over seconds squared

Once you take the square root you get meters over seconds m/s or velocity


The second part delta velocity/acceleration or
(v0-v1)/a

(m/s)/(m/s^2) = (m/s)*(s^2/m)

Meters cancel out and you have s^2/s which = s

Pay attention to your units

and plug the numbers in

Answer 7

Use the acceleration=change in velocity/time taken and use this to work out speed=distance /time. I think that should work.

Answer 8

that stuff is basic professor walk easy for them super riddcule hard for you ,luckly im a professor so im gonna help you 500 mph 1500 m meaning (1.5m/s2 ) m/s3

Answer 9

s = 1/2 a t^2

1600 = 1/2 * 1.6 * t^2

=> 44.72 sec

I hope it helps.