Calculate the relative speed of a vehicle relative to the observer.?
918281
Let's assume your range of vision was 0.2 miles infront of you, and 0.1 miles behind with the mirror.  (I'm pulling these numbers out of my behind.)\n\nAt t=0, the car was at position x = -0.1 relative to you.\nAt t=5 seconds, it was at x = 0.2.\n\nSo the speed (relative to you) is 0.3 miles / 5 seconds = 0.06 miles per second = 216 miles per hour.  Since you were doing 85 mph, it real speed was 301 mph.\n\nAre you sure this took only 5 seconds?  I picked some pretty small numbers for the range of vision and this still results in some wicked speed.
400952
Use 1/s+1/s'=1/f  where f=R/2, s is object distance (in this case negative) where the image of the first mirror is the object of the second.  s' is the image distance you are solving for.  I'll leave it for you to work out the details of your homework, but this should get you started in the right direction.
109202
Fnet= MA so you acceleration is 36km/h, so convert that to m/s, which is 10 m/s. To get acceleration, we know that the car started out at 18km/s, which is 5m/s. So the acceleartion = Change of velocity/Change of time. So that comes out 1 m/s/s. So there is 1000 Newtons on the car. The other guy's answer is wrong. I'm an engineering major. The standard Units in science is MKS, meters kilograms and seconds
788625
This is a momentum question.  If the end result is comming to a halt, then then each vehicles' momentum was equal and opposite.  Momentun (M) = mass (m) * velocity (v),  therefore you are left needing to solve the equation m1*v1 = m2*v2...you can do the rest I hope....
210248
Top Speed 330.0 kph / 205.1 mph \n0 - 62 mph 3.8 seconds
841492
a)The acceleration a = Force / mass = -6070 / 2600 = -2.33 m/s^2\nInitial velocity u = 34.9 m/s\nThe velocity at any time is given by v = u+a*t = 34.9 - (2.33*1.76) = 30.8 m/s\nb) v^2 - u^2 = 2*a*s, where s is the distance travelled in that time. \ns = ( v^2 -u^2)/2a = 57.8 m\nc) The car stops when u + a*t = 0.\nt = 34.9/2.33 = 15 s
1009103
We have a formula:\nSpeed = Distance / Time\nhere Time= Distance /Speed.==D/S\nD=100km in both cases\nS1= 125km/hr; S2= 80km/hr\n\nT1=D/S1==100/125 hrs === 4/5hrs===4/5*60mins= 48mins\nT2=D/S2==100/80 hrs ===5/4 hrs ===5/4*60 mins= 75mins\n\nTime saved = T2-T1==75-48=27 mins.\n\nIs this clear buddy
671239
I have no idea what your question is asking, but lets try this.\n\nbool speeding_or_not(int driver_speed, int speed_limit) {\nreturn (speed_limit < driver_speed);\n}
532915
speed=distance/time\nso ans=118.5/14 =8.4642km/hr
200117
distance= rate*time\ntime= distance/rate
76533
seeing an object depends on reaction time of the eye (usually in tenths of sec). Considering the car is moving with uniform velocity then reaction time can be roughly estimated as\n\nt = d/v , where d is distance between you and car and v is velocity of car (here same as light). \n\nso the answer to your question depends on 'd'
946074
avg. speed = distance/time = 75/11 = 6.82 m/s
684454
Let's consider the case when the Car is right in front.\n\nAt that instance, the car is travelling tangentially to a circle of radius 132ft, where the Camera is the centre of the circle.\n\nCircumference of the Circle = 2*(pi)*R\n= 2 * pi * 132 = 264*(pi) ft.\n\nTime taken to travel the entire circumference by the car\nT = circumference/velocity = 264*(pi)/264 = pi seconds\n\nHence the car would require pi s to cover 360 degrees of the circle.360 degrees = 2(pi) radians.\n\nW = angular velocity = 2(pi)/pi = 2 radians/sec.\n\ncan't help about half a sec later now
702361
ok biker bike, here we go.\n\nacc.= g = 9.8 m/s2, (any object moving in air will have this acc. or retardation)\n\nb) we apply formula\ns=ut + 1/2at^2\nu=20\nt=4\ns=20*4 + 1/2 *9.8 *4*4\n  =158.4 m\n\nc) v=u + at\n  therefore, t= v-u/a\n                   =50-20/9.8\n                   =3 seconds (approx)
973163
Relative to a woman in the car:\n\n(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.\n\n(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.\n\n(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.\n\n(d) a, acceleration of the point at the top of the tire is given by:\n\na = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2\n\n(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction\n\n(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2\n\nNow relative to a hitch hiker on the ground:\n\n(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.\n\n(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.\n\n(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.\n\nω = v/r = (22.22)/(0.33) = 67.33 rad/sec\nv(top) = ωd = (67.33)(0.66) = 44.44 m/s\n\n(j) a at the top of the tire is = v(top)^2/d\n    = (44.44)^2/(0.66) = 2,992.29 m/s^2  \n\n(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.\n\n(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.
437661
This requires the use of doppler shift formula for a moving source and stationary observer:\n\nf = f0 ( v / ( v + u ) )\n\nf0 = actual frequency = 5\nv = speed of wave = 20\nu = relative velocity of source away from observer = -10\n(since travelling toward)\n\nf = observed frequency = 10 Hz\n\nNow velocity = v = fl = frequency * wavelength\n\nso wavlength l = v / f = 20 / 10 = 2m
56541
You would be in a different universe than you are now, because the laws of our universe say that a massive object can't go faster than the speed of light (or even as fast).\n\nAlso, if you changed the question to "...1 mile per hour less than the speed of light", you would have to specify "relative to ..." (what?)  And then you would have to say, what would happen as observed by (whom?).  In this question, relative to you, the headlights would turn on and you would be able to see things 186,000 miles ahead of you (as you measure miles) within 1 second (as you measured seconds) .  Relative to the observer that thought you were moving 1 mile per hour less than the speed of light - assuming you were moving toward the observer - you would get to him or her just a tiny fraction of a second after he/she saw the headlights turned on.  Your measurement of the speed of light would be consistent, but your measurments of time and distance would be very different.