what distance is needed by a car to stop if her spee3d is 60 km/h?
105021
Depends on the rate of deceleration of the vehicle (as well as surface characteristics).  However a rule of thumb they teach in driver's education is one car length for every ten mph of speed.  So at 36 mph/60 kph, that would be about 4 car lengths, or about 20 meters (unless you drive a mini).
1118371
double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t
685470
the speed = distance/time
832829
Speed = distance/time\n\nlet x be the time it takes to get to school on time\n\n3=d/(x+15/60)                                      4=d/(x-15/60)\nd=3x+3/4                                         d=4x-1\n\ntherefore 3x+3/4=4x-1\nx=1  3/4 = 7/4\nHence distance=d=3(7/4) + 3/4 = 6\nTherefore the school is 6 km away
1184449
It does not matter how many seconds your car moves in first gear, after that it should be 60 miles/hour and the feets you do the math by yourself. GOT IT???DUHHH
904741
The change in momentum of the automobile is equal to the impulse of the applied force.\n\nInitial momentum = mv = 1200*50 = 60000 km-kg/h\nFinal momentum = 0\n\nWe need to convert the 60000 km-kg/h to useful units\n\n60000 km-kg/h * 1000 m/km / 3600 sec/h = 16667 m-kg/sec\n\nImpulse = force * time = F*0.14 sec\n\nSo F*0.14 = 16667\n\nForce F = 16667/0.14 = -120 KNewtons\n\nNote that the applied force is against the original direction of motion, so the force is negative...
642636
All that is needed is the MPH must be divided by 60. MPH/60=MINUTES
715166
Distance is found to be the product of velocity and time.  E.G. if you travel at 60 MPH for 2 hours, you travelled (60 M/h) * (2 h).  The hours units cancel giving the answer as 120 miles.\n\nUsing this formula:\nd = v * t\n57 = v * 1.8, so v = 57/1.8 = 31 2/3 FPS.\nTo convert FPS to MPH,\n(31 2/3) * (60*60/5280) = 21.6 MPH.
786122
Acceleration is defined as the change in velocity (speed) with respect to time. In your problem statement, the phrase "slowing down uniformly" implies that the change in velocity is happening at a constant rate.\n\nKnowing this, you can compute the acceleleration as:\n\na = (v2 - v1) / t\n\na = (5 - 20 m/s) / (4 s)\n\na = -3.75 m/s²\n\nUNDERSTAND THIS: The negative sign means that the velocity is slowing down. It does not imply that the velocity is negative.\n\nFor part (a), force is defined as the product of mass and acceleration. In other words, a force of 1 N is required to acclerate a 1 kg mass at a rate of 1 m/s². Mathematically:\n\nF = m * a\n\nYou're given the mass of the car, and you now know its acceleration, so the average force acting on the car is simply:\n\nF = m * a\n\nF = (2000 kg) * (-3.75 m/s²)\n\nF = -7500 N\n\nThe negative sign means that the force is acting against the car's direction of travel.\n\nFor part (b), the distance the car travels in this time is calculated by the distance equation:\n\nS = (1/2) * a * t² + vi * t + Si\n\nWhere vi is the car's initial speed and Si is the car's initial displacement (distance). Initially, at t=0 s, we'll assume that the initial displacement is 0 (because we only want to know how far the car travels from when we start the clock at t=0). Additionally, vi is given as 20 m/s. Plugging everything into the distance equation provides:\n\nS = (1/2) * a * t² + vi * t + Si\n\nS = (1/2) * (-3.75 m/s²) * (4 s)² + (20 m/s) * (4 s) + 0\n\nS = -30 m + 80 m\n\nS = 50 m\n\nThe car travels 50 m during this period of slowing down.\n\nHope that makes it all clear! :)
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
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.
830021
35(2/3) - 39(4/7)\n(70/3) - (156/7) = 22/21\n\nANSWER\n22/21 miles
100012
A would take 22 minutes to get to 15 miles and 32 minutes to get to 20 miles. Unfortunately, there's insufficient information to be precise on B's journey. Did he spin off before the start?.. or at which point en route? Guessing it, I would say A arrived at 15 miles and at the 20 miles point first. NB. I have had to make certain assumptions in your question to arrive at this answer (eg I asumed A slows down to 40mph after 4 miles?)
205365
Let r = the rate on the country roads.  Then r-9 is the freeway rate since he travels 9 mph slower there.  The formula needed is d=rt, or in another form t = d/r where t = time, d = distance and r = rate.  Since both have the same t, \n\n20/r = 15 (r-9)\n\ncross multiply to get 20r - 180 = 15r\n\nadd 180 to both sides and subtract 15r from both sides to get\n\n5r = 180   Divide by 5 to get\n\nr = 36.  So he goes 36 mph on the side roads.
761121
She went 16 miles in 2/3 hrs, so she is going\n(16)/ (2/3) mph=24 mph\nAt 24 mph she will go 30 miles in 1 1/4  hrs or 75 minutes.\nShe needs to keep going for another 35 minutes.
782037
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj