A few acceleration problems ?

Question

1.What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is r = 2.76 0.08 m?

(i know uncertainty but not for a spherical ball!)

would it involve 4/3(pie)r^3?

2.A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 60.0 m high.

How much later does it reach the bottom of the cliff? What is its speed just before hitting? What total distance did it travel?

(i understand im asking a lot but i know some but im just not getting all these above trig step! let it be noted, i did buy a calc book and im going to pass this class.)

3.A car traveling at 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.76 m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g = 9.80 m/s2

( this one i thought i knew but i guess not.)

(95km/h)x(1000m)/(3600s)X(.76)

is that not what im post to get? -35g/s is my answer.

once again.

i really just want an understanding on how to solve these.

thanks for any help offered.

Answer 1

My Try.

2nd Question:

When the stone is thrown up.

Initial Speed = 12m/s Final Speed = 0

A = -9.8m/s*s

Using v = u + at

t = 1.22s

Using S = ut + (1/2)(at*t)

S = 7.35m

Now height of ball = 60 + 7.35 = 67.35m

a=  9.8m/s*s

Initial speed = 0

Using v*v = u*u + 2as

v = 36.33m/s

Using v = u + at

t = 3.7s

Total Time = 1.22 + 3.7 = 4.92s

Total Distance = 7.35 + 67.35 = 74.7m

Final Speed = 36.33m/s

:)

Answer 2

1. formula:  (true value- experimental value)/true value

2. Vf down=sq.rt.(Vo^2+2gh)

    

     t=time up+ time down

     t up= -Vo/g

     t down= Vf/g

     g=acceleration due to gravity= 9.8m/s^2

     Vf in the upward motion is equal to zero, Vo=12m/s.

     Vf in downward motion could be gotten from the first equation given, Vo is equal to zero.

    

    Stotal=Sup+Sdown

    Sup=-1/2gt^2

    Sdown=1/2gt^2

3. You should multiply .76 by 2. And square the value of Vi, that is 696.37 m/s. Then divide: (Vi^2)/2S. Then get the square root. That will give you an acceleration of 21.404m/s2.

Divide this value by 9.8 to give 2.184 g's as the final answer.

P.S.

The answer of the person below is ultimately wrong, the total distance traveled is equal to the distances in the upward and downward motions, not the height of the cliff plus the upward motion's distance. See for yourself.

Answer 3

q1. Er I don't really understand, sorry

q2.Let's find the final velocity. Take down as positive.

v²=u²+2as

v²=(-12)²+2 x (9.8) x (60)

v²=144+ (19.6) x (60)

v²=144+1176

v=36.3m/s

For total distance we need to work out the height above the cliff top

first work out time to get to its max height

t=(v-u)/a

t=(0-(-12))/9.8

t=1.22s

The distance is then found using

s=ut +½at²

s=-12x1.22 + ½ x (9.8) x 1.22²

s=-14.64 + 7.29

s= -7.35m - This means 7.35m above cliff top

Total distance is 7.35 up +7.35 down + 60 for the cliff = 74.7m

total time can be found using

t=(v-u)/a

t=(36.3-(-12))/9.8

t=4.93s

That was a lot of work!!

q3.

v²=u²+2as

u=95km/h=26.4m/s

0=26.4²+ (2a x 0.76)

a=-697/1.52

a=-458.6 m/s²

a=-458.6/9.8 = - 46.8 g

Hope that helps.