Danny From Poland, joined Apr 2002, 3528 posts, RR: 3
Reply 1, posted (15 years 1 month 1 week 6 days 20 hours ago) and read 742 times:

Okay, the forst question was "the velocity of the car relative to the bus". Your answer was 30km/h and as far as i know, that is correct...

The second question was "the minimium time to pass the bus, given that the bus is 15 meters long"
- Let's pretend that the bus stands still, and that the car then is travelling at 30km/h.
- 30km/h = 30.000m/h = 8.3m/s.
- The bus is 15 m long - 15/8.3 = 1.8s.

Answer: the time to pass the bus is approx 1.8 seconds.

Now, I can be wrong...math wasn't my best subject...

N949WP From Hong Kong, joined Feb 2000, 1437 posts, RR: 1
Reply 3, posted (15 years 1 month 1 week 6 days 4 hours ago) and read 721 times:

Don't forget the length of your own car!

As you begin to pull alongside the bus to pass it, the front of your car will first line up with the rear of the bus and gradually move forward until it lines up with the front of the bus. This will take 1.8 seconds, but your car will still be alongside the bus -- ie. you still haven't completely passed it yet. Divide the length of your car (in meters) by the 8.3 m/s will give you the additional time required for your car to completely pass the bus.

Pandora From , joined Dec 1969, posts, RR:
Reply 4, posted (15 years 1 month 1 week 5 days 19 hours ago) and read 717 times:

Damn i was wrong at the first place. My physics teacher said it could be a simultaneous equation and i did too initially too!!! But it was easy when danny explained it in a very good way. thanks alot.

and also another question which is also very very tricky too.

here it is:

a woman of mass 60kg is standing on a set of scales in an elevator. Calculate the reading of the scales when:

a. the lift is stationary. i know this one, it is still 60kg.

b. the lift is accelerating upwards at a constant at 2.0m/s

c. the lift is moving upwards at a constant speed.

d. the lift is moving upwards and decelerating at 3.0m/s.

b and c and d are very hard. i reckon it's soemthing to do with newton's 3rd law and one of the straight line motion equations or f=ma.

StarAlliance From Hong Kong, joined Dec 1999, 252 posts, RR: 0
Reply 5, posted (15 years 1 month 1 week 5 days 16 hours ago) and read 713 times:

Pandora, it is actually a really easy question!

But, first of all, I think you have typed it wrongly that acceleration is 2m per second square, NOT PER SECOND. Check it out in your book!!

Regarding the answers:
b) When the lift accelerates upwards, a net upward force is required to accelerate the lift. The force R exerted by the lift floor is greater than our weight MG, right? Therefore, the net force is R-mg=ma

Thus, weight of the women=mg=60kg x 10 m per second square=600N
Then,
R-mg=ma
R-600N=60kg x 2m per second square
R=720N
Therefore, the reading of the scale is 72kg!!!

C)much more easy, The reading = the mass of women = 60kg!!!

d.) As I have mentioned in B, it have a similar equation mg-R=ma Since when the lift accelerates downwards, a net downward force is required. The force R exerted by the lift floor is less than our weight mg.

R-600N=70kg x -3m per second square
R=390N
Therefore the reading will be 39kg!!

That's easy!

Hopes it can help!

PS Actually, I am a grade10 students in Hong Kong, ( a city of China ) and your question help me revise the stuff!! Thanks anyway. When I compared your question to my question here, I found that my question is actually 10 times more harder!!

XNV From Canada, joined Jan 2000, 142 posts, RR: 0
Reply 6, posted (15 years 1 month 1 week 5 days 12 hours ago) and read 704 times:

If you want an easy way to remember how to convert m/s to km/h and vice-versa, remember this:

to go from m/s to km/h multiply by 3.6 (think km are larger than m so you multiply).
to go from km/h to m/s divide by 3.6 (think m are smaller than km so divide).

On an exam you may have to show your work how you got there but this is a quick way to check to make sure you are right.