Please this is a very important physics question to me?

Question

A car that is leaking oil at a constant rate is uniformly accelerated from rest starting out just when a drop falls. The first drop down the road is 0.80m from the car's starting point. How far from the starting point is the 4th drop?

Answer 1

d1 = ½at²

t = √(2d/a)

4t = 4√(2d/a)

d2 = ½a16(2d1/a) = 16d1 = 12.8 m

Or even simpler, distance is proportional to time squared.

4² = 16, distance is 0.8*16 = 12.8 m

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Answer 2

0.8 + 8.8=9.6 m.

By the time the first drop of oil has leaked the car has already in motion and has covered 0.8 m, for the first stage the initial velocity is 0.In the second case you have three more drops to fall when the initial velocity is 1.6 m/s and you are left with three more drops to fall.Assuming the interval between two drops leaking you are left with three more drops to leak and thus the calculation.

Answer 3

[EDIT: There's an ambiguity in the meaning of the term, "the 4th drop."  It could mean either:

1. "the 4th drop 'down the road'" (i.e., the 3rd drop following the 0.80m drop); or:

2. "the 4th drop that came out of the car" (considering that the 1st such drop came out at the starting position).

Most responders are assuming interpretation #1.  I'm taking the question at face value and assuming interpretation #2.  You get different answers depending on which interpretation you choose.]

The oil drops act like a ticking clock; you can count them to measure how much time has passed.  Specifically, you can use the following formula:

If:

n = number of drops that have dripped;

Δt = amount of time between drips (which we don't know)

t = total time that has passed;

Then:

t = (n-1)Δt

(plug in a few values of "n" to convince yourself that that formula is correct)

Now combine that with the formula that relates time, acceleration and distance:

d = ½at² (starting from rest)

In terms of oil drops this is:

d = ½a[(n-1)Δt]²

Now, the "first drop down the road" is actually the 2nd drop (because the problem says the _first_ drop is actually at the starting point).  So substitute "n=2" into the formula.  And call the distance "d2" (where "2" stands for the drop number).

d2 = 0.80m = ½a[(2-1)Δt]² = ½aΔt²

Let d4 be the distance traveled from the starting point to the 4th drop:

d4 = ½a[(4-1)Δt]² = ½a(9)Δt²

Now, notice that if we divide d4 by d2, we can get some annoying variables to cancel out:

d4/d2 = ½a(9)Δt² / (½aΔt²) = 9

So now we're left with "d4/d2 = 9".  The problem tells you that d2 = 0.80m, so now it's a simple matter to solve for d4.

Answer 4

a-is acceleration

t-is time between  drops

4*t -is the time passed between 0-th and 4-th drops

x -is the distance to the 4-th drop

do you remember this formula : distance=1/2*a*t^2

0.8=1/2*a*t^2

x=1/2*a*(4*t)^2=(1/2*a*t^2)*16

x=0.8*16=12.8 m

i hope you understood.

Answer 5

You need to have some more information. Suppose that the drops fall every second.

Therfore you can say that the car traveled 0.8m in the first second

Calculate the acceleration of the car:

s = ut + ½*a*t²

0.8 = 0 + ½ *a* 1²

a = 1.6m/s²

The 4th drop will have fallen after 4 seconds:

s = ut +½*a*t²

s = 0+ ½*1.6*4²

s = 0.8*16

s = 12.8m.

The 4th drop will fall 12.8 metres away IF THE FREQUENCY OF THE DROP FALLING IS 1/s. This is not given and is an assumption.

Answer 6

Let time interval between the dropping of 2 drops of oil be t.

For 1st drop after the starting of d car---

time=t

s=0.8m

u=0

acceleration=a

0.8=(a*t^2)/2

a*t^2=1.6

a=1.6/t^2

For 4th drop---

time=4t

acceleration=a

u=0

dist=s

s=(a*16*t^2)/2

s=25.6/2=12.8 mts.