Danny From Poland, joined Apr 2002, 3509 posts, RR: 2
Reply 1, posted (14 years 3 months 9 hours ago) and read 655 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 (14 years 2 months 4 weeks 1 day 17 hours ago) and read 634 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.
StarAlliance From Hong Kong, joined Dec 1999, 252 posts, RR: 0
Reply 5, posted (14 years 2 months 4 weeks 1 day 5 hours ago) and read 626 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
R-600N=60kg x 2m per second square
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
Therefore the reading will be 39kg!!
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!!