a motorbike going at 80m/h and a big bus going at the same speed, which arrives first?
969523
if they are both going at the same speed, then the mass,size of the wheel and other considerations have already been take in account, so they both should arrive at the same time.\nGod bless,\ngabe
40969
Buy Motorbike its traffic-free ... lol
125621
B) Increase\n\nThe difference in the two objects' velocities would be constant since that are both accelerating at the same rate, but the first object already had a higher initial velocity (from Object B' point of view).\n\nThe equation relating the distance traveled with respect to initial velocity, accelertion and time is,\nd = v_i * t + 1/2 a*t^2\nSo for both objects, after B is released the 1/2 a*t^2 part of the equation is the same, so we can ignore that part and focus on the v_i * t part.\nIf we start t=0 at the moment object B is released, object A's velocity is some non-zero number while B has no initial velocity.  So as time goes on, the distance between the 2 objects inceases.
668935
it would be the exact same speed!  \nhere is an example. --------\n--if you are swerving on the road and going 50 mph. and then you stopped swerving, you would still be going 50 mph
950795
the answer all lies in the buses wheel. if its speedometer is the same as for a small vehicle. then you are right the bus is alot faster. you see when an engine of a bus makes the same number of rotations as for a small car tyhe distance covered is greater since the tires are large thus the curcumfrence is larger. principle applied:\n\nspeed =distance*time,\ndistance =curcumfrence of tyre \nno of rotations may be the same But. if the bus tyre say be 5meters curcumfrence .\nthen for the small car is 2meters curcumfrence.\nthe mo of rotations on both cars are 10 per second \n\nif the speedometer records its speed using the no of rotations then. the speed of both vehicles will be the no of rotations (in the whole curcumfrence of the tires)*time so \n(5meters*20rotations)/2 seconds=50meters per second\nwhile as for the small car \n(2 meters*20rotations)/2=20meters per second\nif the speedometer records the no of rotations on the axis the definatley the bus covers more ground in less time. thus the effect of feeling moving faster than a small car while traveling in the same speed.
456275
9.81 m/s² or 32.2 ft/s²
792509
At 1 p.m the 1st bus starts moving west with a velocity of 44 mph  that means at 2 pm the bus will be 44 miles away from the station because  D = V x t = 44 mph x 1 h = 44 miles .\nAt 2 pm the 2nd bus starts moving east i.e in the opposite direction with a velocity of 48 mph \n              \nB1  ____________B2      ( B1 B2 = 44 miles )\n                      \n             \n\nAfter 2 pm the distance between the 2 buses at any time is equal \nto : \nD = 44 +( V1 x t ) + (V2 x t ) = 44 + (44 x t ) + (48 x t ) = 44 + 92 t \nwhere D >> miles , t >> hrs \n                                         \n                                          B1    _________S___________ B2\nWhere B1 S = 44 miles + 44 t \n            B2 S = 48 t \n                                             \nPut D = 274 miles to get  t \n274 = 44 + 92 t     >>  t = 2.5 h\nTherefore the two buses will be a distance of 274 miles apart at : \n2 pm + 2.5 h =  4 : 30 pm
65812
It is currently 763 mph (or mach 1.02)
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
530622
Ok, first you want to see where they meet. To make this easier, advance train A for 5/6ths of an hour and see the problem in terms of when train B starts out. 65 x 5/6= 325/6\nkilometers. So the problem is the same only they start out at the same time over a distance of 600- 325/6 kilometer =\n545 5/6 kilometers.\n\nNow, they travel in the proportion of 65/105 kilometers per hour. So the distance till they meet will be in the same proportion. That is 545 5/6 x 65/105 = 212875/630 or about 338 kilometers for train A to travel. Since it travels at 65 kph, this means it travels for 5.2 hours from the new start time equal to B's which was 6:20 AM. Ends up being 11:32AM.
73531
As above. The speed is dependent on the angle and length of the ramp and the level of air resistance (i.e. if you're riding into the wind). However, the mass of the bike is irrelevant, a heavy object would follow the same path as a light object. Roughly speaking.
192470
About 85mph before the factory restrictor kicks in.
816469
well F=mv/t given: m=78kg, v=24m/s, t=60s\n\nF=78(24)/60=31.2N
3944
The Trolley + Child being one single system, the speed of the Centre of Mass would not vary. It would remain the same velocity V.
933881
t1=s1/v is time before yodeling, s1=20 miles, v=168mph; s2=t2*v2 is path of yodeling, t2=3/60 hours, v2=98mph; s3=t3*v is path of the bus after yodeling, t=t1+t2+t3=1 hour or t3=t-t1-t2, s=s1+s2+s3 is total path of silent+yodeling+shut_up phenomena; thus u=s/t is average speed of the bus;\nu=(20 + (3/60)*98 + (1-20/168-3/60)*168) / 1=\n=164.5mph
677621
It's 220mph, near enough, so you're looking at perhaps a bullet, or a jet. It might be worth dropping it to 49 m/s (110mph) or even 24.5 m/s (55mph) to bring it within the range of a sports car or ordinary motor.
947159
Assuming the first bus goes west it would be 4:30 PM.