physics question?

Question

1.) a golf cart has an acceleration of 0.4 m/s square. what is its velocity after it has covered 10 m starting from rest?

(answer: 2.8 m/s)

how do u get the answer.

Answer 1

v^2 = u^2 + 2 a s

a = 0.4, u=0 , s =10

so v^2 = 2 * 0.4 * 10 = 8

so v = 2.8

note: derive the formula used here... start with v=u+at, s=ut+1/2 at^2, ...

Answer 2

s(t)=(1/2)at^2+(v0)t+s0

In this case, the starting position s0 = 0 (arbitrary, but otherwise s(t) will be 10+s0), v0= 0 ("from rest"), a = 0.4, and s(t) = 10.

10=(1/2)(0.4)t^2
10= (0.2)t^2
t^2 = 10/(0.2) = 50
t = sqrt(50)

now v = at

v= (0.4)sqrt(50)

Unfortunately, proprer notation cannot be shown. The limits of Y!A made it unreadable to include the units in the calculation. Do not forget to keep the units with the values. v0 and s0 should be subscripted. Follow the notations and formats that you are expected to in your class to get full credit.

Answer 3

You need the equations of motion (see link).

Using: v^2 = u^2 + 2as

where v^2=final velocity squared
u^2=initial velocity squared
a=acceleration
s=distance travelled

Now, we start at rest so u=0 and u^2=0
We know a=0.4 and s=10, thus:

v^2=0+(2x0.4x10)
v^2=8
v=squareroot (8)
v=2.828... m/s
And we have our answer

Answer 4

hello
this is a kinematics question.
use the following equation: v(square)=u(square) + (2 x a x s)
let me define the variables;

'v' stands for final speed
'u' stands for initial speed
'a' stands for acceleration and
's' stands for distance

since the object starts from rest, the initial speed will be zero. it is the final speed which you have to find out.
so replace the values in the equation:

v(square)=u(square) + (2 x a x s)

v(square)= (0)(square) + (2 x 0.4 x 10)

v(square)= 0 + 8

v = square root of 8
= 2.83 m per second

which is approximately 2.8m per second
hope you understood

Answer 5

s = 1/2at^2

where s is distance, a is acceleration, and t is time.

10 = 1/2 (0.4) t^2 ==> t = sqrt(20/0.4) = sqrt(50) = 5 sqrt(2) seconds

Next formula: v = at = 0.4 (5 sqrt 2) = 2 sqrt 2 = 2.8 m/s (Answer)

Answer 6

v2^2-v1^2=2a(Delta)d where v1 is its starting speed (0m/s) and v2 is its final speed. ^2 means squared. Thus, solve for v2
a=0.4m/s squared
v1= 0
delta d=10m
v2^2 - v1^2=2a(delta)d
v2^2 - (0) = 2(0.4)(10)
v2^2=2(4)
v2^2=8
take the square root of 8
v2=2.82

Answer 7

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

If u = 0 starts from rest

Rearrange (1) gives:
t = (2s/a)^0.5
Substituting in (2) gives:
v = (2sa)^0.5

s=10m a=0.4ms^-2
v = 8^0.5 = 2.828ms^-1

Answer 8

(v-v0)^2 = 2as
v = v0 + (2as)^(1/2)

is derived from the differential equations

s = dv/dt

and

a = dv/dt

Answer 9

put the peddle to the metal and use a stop watch.

Answer 10

I would post the question on answers.yahoo.com and let someone else do the work