how to do acceleration and velocity problems!? please help!!!! asap?

Question

i tried doing these problems but my teacher said i got them wrong and to re do them.. can someone help?

a stone is dropped from a cliff. after it has fallen 40 m, what is the stone's velocity?


and also

what is the average acceleration of a car tha tgoes from rest to 59 km/h in 9.0 seconds?



and the last one


a ball is thrown up with a velocity of 60 m/s
at what time will it have a velocity of 0 m/s?

ahh please help! i greatly appreciate it

Answer 1

1)d=40m
in things which are in free falling state or motion you will use the acceleration which is 9.8m/s^2(gravitational acceleration;represented by g) so with the given data

v2^2=v1^2+2Gd

v1 is equal to zero because it started from a high level and dropped down or it started at rest.
and v2 is what you are looking for so

v2^2=2Gd

substitute the values

v2^2=(2)(9.8m/s^2)(40m)
v2^2=784m^2/s^2

extract square root at both sides

v2=28m/s

so the velocity after it fall the 40m height is 28m/s

2)a=v2-v1/t

the given have different values time so convert 59km/h to m/s(usual) so

59km/h=16.39m/s(it is hard if i will show the whole step so just solve it yourself and you will get the same answer)
then substitute the values

a=(16.39m/s-0)/9s(the vi is zero because you said it started from rest)

then

a=1.82m/s^2(the average acceleration that you are asking)

3)because it is still an object in free fall(although it is inverse of the 1st one) so acceleration will still be 9.8(same unit) but it will become negative(-) because it moves upward and the velocity is decreasing so

the equation will be

-9.8m^2/s^2=(0-60m/s)/t (because you are finding for time)

t=(-60m/s)/-9.8m^2/s^2

so t=6.12s is the answer to your question

hope this helped....

Answer 2

There is a -reason- that your Physics book has all those funny looking equations in it. You use them to solve problems such as these.
Somewhere in your book you'll see an equation that says
v² = v0² + 2ax. That means that the square of the final velocity is equal to the square of the initial velocity plus 2 times the acceleration times the distance travelled. So..... For the 1'st problem v0 is zero, the acceleration is 9.8 m/s² (Earth gravity) and the distance is 40 meters. Then
v² = 2*9.8*40 = 784 and taking the square root of both sides,
v = 28 m/s.

Now -you- dig into your book and figure out how to do the rest of those problems. I learned this stuff 50 + years ago and if I do it for you, then you won't learn anything ☺

Doug

Answer 3

Draw a diagram!!!!!

Draw a diagram showing starting position, end position, any velocities, accelerations and times.

Take one direction to be positive. In the last question, the initial velocity is up, but the acceleration is down. Therefore they must have opposite signs, or the maths will not work.

Then, use these equations:

(final velocity)^2 = (initial velocity)^2 + 2 * acceleration * displacement
v^2 = u^2 + 2as

v = u + at
s = ut + 0.5 a t^2
s = 0.5(u + v)t

Remember, these equations only work for constant acceleration (eg, only gravity is affecting the object)

These are the answers, but your teacher will expect the working to be shown:

1. 28.0m/s
2. 6.56km/h/h
3. 6.12s

Answer 4

the respond is B. The quantum of light is a particle and is consequently localized... aka one component... and might in basic terms deposit the skill at a component, in spite of the reality that the region that component could be is randomly allotted in accordance to 3 threat density function which has an intensity proportional to the intensity of the mild on the show reveal integrated over an prolonged term.

Answer 5

1)
s = 40 m
the initial velocity = 0
g = 9.86 m/s^2
v^2 = 2gs
v = sqrt(2gs) = 28.1 m/s
2)
v = 59km/h = 16.4m/s
v = v0 + at
v0 = 0
a = v/t = 1.8m/s^2
3)
a = g = -9.86m/s^2
v = v0 + at
v0 = 60m/s
v =0 m/s
t = 60/9.86 = 0.6 s