physics questions, NEED HELP!!!!?

Question

Please help me with these questions and please explain how you got your answers!!! Thanks

1. A honda moving at 100 km/h is 80 km behind a Ford moving in the same direction at 60 km/h. How far does the Honda have to travel before it catches up to the Ford?

2.A spacecraft takes 200 km after being launched to reach the escape speed fro the earth of 11.2 km/s. Find its average accleration in terms of g.

3. A car that is leaking oil at a constant rate is uniformly accelerated from rest starting out just when a drop falls. The frist drop down the road is 0.80 m from the cars starting point. how far from the starting point is the fourth drop?

4. A bus travels 400 m between two stops. It starts from rest and accelerates at 1.5 m/s until is reaches a velocity of 9.0 m/s. The bus continues at this velocity and the decelerates at 2.0 m/s until it comes to a hault. Find the toal time required for the journey.

Answer 1

I'll start you off, in question 1
the honda is travelling 40 km/hr faster than the ford. Since they are separated by 80 km, it will take the honda 2 hrs to catch up.
At 100 km/hr this will be 200 km travelled by the honda.

Answer 2

The honda's position can be described with the equation: H=100t
The ford's position can be described with the equation F=60t + 80

Setting the two equations equal will allow you to find how much time it will take for the ford to catch the honda.

100t = 60t + 80
40t = 80
t = 2 hours

Then put the 2 hrs into the equation that tells where the honda is: H=100(2) = 200 km.

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You should have an equation like this:

Vf^2 - Vo^2 = 2ad
11.2^2 - 0 = 2 a (200)
a=.314 km/s^2 = 314 m/s^s

g = 9.8 m/s^2, so this rocket's acceleration is 32 g. (Unfortunately, this would kill any astronauts on board.)

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You should have an equation like this:
Xf = (1/2)at^2 + Vo t + Xo

0.8 = (1/2)Acar (1drop)^2
D= (1/2) Acar (4 drops)^2

Plug in D1 and T1 for the first drop. (Here we will define a new unit of time, "the drop".) Then use the same equation to consider the distance the car will travel in 4 drops.

Now, you have two equations and 2 unknowns. Solve the top equation for Acar then substitute that into the bottom equation and solve for D. (I get 12.8 m)

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You should have an equation like this:
Vf = at + Vo
9 = 1.5 t + 0
t = 6 sec are spent accelerating.

In that time the bus travels 9^2 - 0 = 2(1.5) d
d= 27 m

Then while decelerating, use the Vf = at + Vo equation,
0 = -2t + 9
t = 4.5 sec

While decelerating, the bus travels 0-9^2=2(-2)d
or 20.25 ms.

So the bus travels 20.25 + 27 = 47.25 m while speeding up and then slowing down. That means it travels 400 - 47.25 = 352.75 m at the constant speed of 9 m/s.

That takes 352.75/9 = 39.19 seconds

So the total time is 39.19 + 4.5 + 6 = 49.69 sec

Answer 3

1)
The ford’s velocity is 60km/h.

The Honda’s velocity is 100km/h.

Both are in the same direction.

Now imagine a wind with a speed of 60km/h begins to act in the opposite direction of both the cars.

Now apply minus 60k/h simultaneously to both of them.

The ford’s velocity is reduced to zero and that of Honda’s speed is reduced to 40 km/h.

The distance between them is 80 km.

To move through a distance of 80 km, the Honda takes
80 /40 = 2 hour.

2)

Initial speed is 0.
Final speed is 11.2 km/s.
Distance moved is 200km.

Average speed is 11.2 /2 = 5.6 km/s

Time taken to travel 200 km is (200 / 5.6) = 35.71 s.

Average acceleration is, ‘change in speed / time’
= 11.2 / 35.71 = 0.31km/s

3)

The problem is not clear.

The car is accelerated uniformly from rest just when a drop falls. Therefore the drop’s initial down ward velocity is zero and also it has no horizontal speed.

How the first drop will be at 0.8m from the start?

Therefore, let us assume that initially a drop is on the floor and the car starts when a drop hits the ground.

Also let us take the drop that drops after this start as first drop.
For the first drop

S = 0.5 a t^2

a = 1.6 /t^2

Fourth ball begins to drop after 3t.

Total time of travel = 4t.

Total distance is = 0.5 x a ( 4t)^2
= 0.5* (1.6/ t^2)*( 4t)^2
=25.6 m

4)
t1 = v/a
t1 = 9/1.5 = 6s.
Distance traveled in 6s = 0.5 x 1.5 x 36 = 27 m

t3 be the time with deceleration of 2m/s^2
t3 = 9/ 2 = 4.5 s.
Distance traveled in 4.5s = 0.5*2*4.5*4.5 = 20.25 m

t2 be the time with uniform speed of 9m/s

Distance traveled with this speed is
400 - 27 -20.25
352.75m

t2 = 352.75 /9 =39.19s

Total time is 39.19 + 6 + 4.5 = 49.7s