one motor driver from calm state start to move and then after going 120m ,its speed reach to 30/0 m/s .indicate motordriver movement acceleration and the time he went this distance .(please write the way of solution)
206095
There is no one answer to this question. You need to know either the rate of acceleration or the time taken to get to 30m/s.\n\nYou will then be able to solve with the formula:\nv = u + 2as\n\nWhere v is the final velocity, u is the initial.\na is the acceleration.\ns is the time.\n\nTherefore:\n30 = 2as\n15 = as\n\nSo as long as acceleration and time equal 15 when multiplied together then that is a valid answer.
314934
Okay, I've thought about this and it's not possible to determine the rate of acceleration from the given criteria.  It is an "average" of 60 miles per second, BUT how many miles in the first second compared to the number of miles in the final, fifth second?  If that is known, you can project MPH from seconds 6 to 10 or the additional 1/4 mile.  How do you calculate the rate of acceleration in your theory?  How fast is the vehicle traveling at each point of the 1/4 mile distance?
554687
velocity = df(t)/dt = 4π + 4πcos(4πt)\nacceleration = dv(t)/dt = -16π^2sin(4πt)\n\nParticle is at rest when velocity = 0, or 4π+4πcos(4πt) = 0\n1+cos(4πt) = 0\ncos(4πt) = -1   The values for which cosø = -1 are ø=n(odd)π, so when\n4πt=n(odd)*π, t = 1/4*n(odd); thus for t = 1/4, 3/4, 5/4, 7/4, 9/4, 11/4 the velocity is zero\n\nThe maximum velocity occurs when cosø = 1, so the max velocity is 8π
27559
distance is speed multiplied by time.\nexample: you travelled for one hour (t=1) and your driving speed was 70mph (s=70)\nThen, D=70m/h x 1h (the units "h" get cancelled out)\nANSWER: 70m\nYou went a distance of 70 miles
510560
The stop distance is not an arithmetic progression (ie it is not 40m). It is in fact a geometric progression. The stop distance is a lot lot more. Truthfully, I don't know but if you were travelling at 120 kmh and there was a truck 40m (or even 60m) in front of you when the brakes locked; you'd be dead.
227949
75(km/s)/Mpc * 120Mpc = 9000km/s.
544629
Split the problem into two parts, first when the car has its tyres still on terra firma and the second it thinks it is Wright brothers reinvented. From the first part we find the velocity of the car just before its leap into thin air. The formula to use for this part will be \nv^2 - u^2 = 2a*s. \nHere u = 0, having started from rest.\na = 4 m/s^2  and s = 50 m. This gives \nv = 20 m/s. Now the second part. This velocity is at an angle of 24 deg. below the horizontal. Its vertical component is 20 sin24 and the horizontal component is 20 cos24. The horizontal component does not change during its flight because the only force acting on the car is vertical. You can get the time of flight using the formula\ns = u*t + 1/2 a*t^2, where s = 30 m; u = 20 sin24 and a = g = 10 m/s^2.\nSolving the quadratic equation for t, you get\nt = 1.78 s approx.\nThe distance from the base of the cliff where the car lands is given by the horizontal velocity * time of flight, that is 20 cos24 * 1.78 = 32.5 m.\n\nAs for the diagram, I do not know how one could attach it as a part of the reply in Yahoo Answers. I have felt this shortcoming in this otherwise excellent forum.
514260
x =70m, t = 7s, Vf = 15m/s\na.)\nx = 1/2(Vi + Vf)t\nVi = initial velocity\nVf = final velocity\n\n70m = 1/2(Vi + 15m/s)(7s)\n10m/s = 1/2Vi + 7.5m/s\n2.5m/s = 1/2Vi\nVi = 5m/s\nb.)\nVf = Vo + at\n15m/s = 5m/s + 7s(a)\n10m/s = 7s(a)\na = 1.43m/s/s\nbasic kinematics\nyou can't use distance = rate x time because there is acceleration and the speed is not constant.
407951
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....
130417
Q1:  4 hours and 15 minutes is 4.25 hours.  The girl's speed is 251/4.25 miles/hour.  Without a calculator, I would estimate that this is 60 miles per hour (60 = 240/4).  With a calculator, it is 59 miles per hour. \n\nQ2.  Martin drives 38 km at 63 km/hr.  He was driving for 38/63 hours.  Without a calculator, I would estimate that this is a little less than 38 minutes (38/60 hours would be 38 minutes).  With a calculator, this is 0.60 hours, which is 36 minutes (0.60 * 60 = 36).
123108
The car accelerated from 0mph to a speed of 100mph.\nIt took 50s to do this.\nThe rate it changed its speed is its acceleration - how long it took to reach that speed.\nSo, 100mph / 50s = 2mph per second.\n\nA car that could accelerate faster reached 100mph in only 25s.\nSo, 100mph / 25s = 4mph per second.
320123
The difference in rates is 1 km/hr. So if they are moving in the same direction, the faster walker will be 5 km ahead at the end of 5 hours.\n\nThe sum of the distances traveled per hour is 11 km. So at the end of 5 hours of travel in opposite directions, they will be 55 km apart.
64315
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
666787
You have mass of car = 1500 kg.\nand speed = V = 25 meters per second.\n\nYou will recall the equation \nE = (1/2) * m * v^2\n\nKinetic Energy = 0.5 * mass * speed *speed.\n\nPunch these numbers into your calculator and you will get a numerical answer with units kilogram meters meters per second per second.\n\n\nThis is the kinetic energy of the car, which is removed by the brakes.\n\nI'm not going to do all your homework for you.  Just the first part to get you rolling.\nHave a great day.
520647
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
199345
speed=distance/time\nV=S/T\nS=VT (distance = speed x time)\n\nlet Vg be gary's speed\n    Vt be Tom's speed\n    S be the distance\n given that Gary completed the run in 20 min, we have \nS=Vg x 20min     (x means times)\ngiven that for time T=20min Tom completed 5/8 of the distance or St=5S/8=Vt x 20min\nsolve St=5S/8=Vt x 20min for S to get S=(8/5)Vt x 20min\n\nnow we have two equations for S, S=Vg x 20min and S=(8/5)Vt x 20min\nit means that Vg x 20min = (8/5)Vt x 20min   /divide both sides by 20min\nVg=(8/5)Vt\n\ngiven that Tom's average speed for the race was 75m/min less than Gary's, we have Vt=Vg-75\nsubstitute for Vt in Vg=(8/5)Vt\nVg=(8/5)(Vg-75)\n5Vg=8Vg-600\n3Vg=600\nVg=200 m/min\nVt=200-75 = 125 m/min (Tom's speed was 125 m/min)\nS=Vgx20min= 200x20=4000m (distance of the race was 4000m)
534206
a) 30*5280/3600 *ft/sec = 44 ft/sec\n\nat 150 feet away it will take you 3.4 secs to reach the child\n\nb) 20*5280/3600 *ft/sec = 29.3 ft/sec\n\nat 150 feet away it will take you 5.1 secs to reach the child\n\nc) 10 miles/hr would give you 15 ft/sec and that is a reasonable reaction time to process the problem and react.  That would give you well over 10 seconds to apply the brakes in an emergency.  Plus it would give you better control over the car.