2 Mechanics Questions

Hi

I am stuck again!!

We were given homework over half term and I finished it apart from the last two :s-smilie:

, So I needed your help. Here are the last two. I am really stuck and don't know where to go at all. Any guidance would be greatly appreciated.

3. (Take g = 10ms^-2 in this question.)
A balloon is moving vertically upwards with a steady speed of 3ms-1. When it reaches a
height of 36 m above the ground and object is released from the balloon. The balloon then
accelerates upwards at a constant rate of 2ms^-2. Find:
a) the greatest height of the object above the ground,
b) the speed of the object as it strikes the ground,
c) the time taken by the object from leaving the balloon to striking the ground,
d) the speed of the balloon when the object strikes the ground.

4. A car is moving with speed u ms^-1. The brakes of the car can produce a constant
retardation of 6ms^-2 but it is known that, when the driver decides to stop, a period of 2/3
second elapses before the breaks are applied. As the car passes the point O, the driver
decides to stop. Find, in terms of u, an expression for the minimum distance of the car from O when the car comes to rest.
The driver is approaching traffic signals and is 95 m away from the signals when the light changes from green to amber. The light remains amber for 3 seconds before changing to red. Show that:
a) when u < 30, the driver can stop before reaching the traffic lights,
b) when 3u > 95, the driver can pass the traffic light before it turns red.

Reply 1

*bobo*

3)a) the object is moving upwards at speed of 3ms^-1 on release and is subject to decelleration of 10ms^-2. use v^2= u^2 +2as. so max height is 36+s
b)conservation of energy
c)use a= (v-u)/t
d)use time found, and a=2, u=3

Reply 2

Glutamic Acid

4) Do (2/3 * u) + (v^2 - u^2)/(2a)

2/3 * u = distance before brakes applied.

Reply 3

The Strangest Quark

3 a) is easy, it's just a one-dimensionsal projectile, with an initial velocity of 3 ms-1, and an initial height of 36m, decelerating at a rate of 10ms-2. You can do this with v^2 = u^2 + 2as , setting v as zero, so that you're finding the distance travelled between being released and reaching its highest point.

3 b) is also easy, you've already worked out the greatest height reached in part a), so now the problem is just that of an object being dropped from that height and accelerating under gravity.

There are a few ways you could do 3 c), I think. the quickest is to use v = u + at, with u being 3ms-1, v being the answer to the last part (remember it needs to be negative as it is travelling in the opposite direction), and a being -10ms-2.

d), again, is very simple using the answers to the previous parts. You know the speed of the balloon, and you also know the time it has been travelling for from part c), so distance is just speed*time.

Reply 4

*bobo*

4) take the two different parts of stopping separately, so how far does it travel at u m/s for 2/3 secs? then using v^2=u^2 +2as for when the brakes have been applied.
a)using the total distance above and adding the distance travelled in 3 secs at u m/s, must be less than 95 in order to stop before traffic lights. solve inequality from there
b)the driver won't have applied his brakes so is driving at constant speed u m/s, and in 3 secs must travel more than 95m in order to pass traffic lights in time