A car(mass=150kg) is moving at a speed of 5.56 m/s how long it takes to raise it's speed to 13.89 m/s in 30 s
115023
Imagine a `speed` x `time` chart. Plotting the data, we will have parallelogram whose area will the length the car walked between the two speeds.\n\nWe'll have, then, 30*((5.56+13.89)/2) = 291.75m
696298
1st way :-\ndistance = speed x time = 55 mph x 3 h = 165 m\n\n2nd way :-\ngiven that d = 55t , so on putting t = 3hrs , we get d = 165 m
860183
This is a momentum question.\n\nMomentum = mass x speed\n\nSo, the original momentum is contained solely withing the moving engine and equals 4 units of mass times 5 km/h.  This equals a 'momentum' of 20.\n\nThe final momentum equals the combined mass of the engine and freight car, which is 4 units plus 1 unit for a total of 5 units of mass.\n\nSo, if Momentum = mass x speed, then since our final momentum is 20 (the same as the original momentum) and our final mass is 5, then 20 = 5 x Speed\n\nDividing both sides by 5 yields a final speed of 4 km/h.\n\nBest wishes and good luck.
716223
The change in velocity of an object undergoing uniform acceleration, a, for a given time, t, is given by:\nv = at\nwhere v is the change in velocity of the object.\n\nIf the object is accelerating at a constant rate of 10 m/s^2 for 5 seconds, the change in velocity the object experiences is 50 m/s.\nSince (we assume) the object started from rest, the objects final velocity after 5 seconds is 50 m/s.\n\nTo find the average velocity, divide the total displacement of the object by the total time.\nIn this case, the object has traveled 125 meters in the 5 seconds it has been accelerating (d = 1/2 at^2), so dividing this by 5 seconds and we get an average velocity of 25 m/s.
532915
speed=distance/time\nso ans=118.5/14 =8.4642km/hr
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
741402
a) F = ma \nYou're given F = -6070 N (negative because it's in the opposite direction) and m = 2600 so a = -2.33 ms^-2. Then over 1.76s, the change in velocity is (-2.33)(1.76) = -4.10 m/s. Car velocity is then 34.9 - 4.10 = 30.8 m/s.\n\nb) v^2 = u^2 + 2as\nv = final velocity = 30.8 m/s (from above)\nu = initial velocity = 34.9 m/s\na = acceleration = -2.33 ms^2 (from above)\nRearrange to get s = (v^2 - u^2)/2a = 155.61 m\n\nc) v = u + at\nYou want v = 0. So if the force is constant then acceleration is constant at 2.33ms^-2. \nu = initial velocity = 34.9 m/s\nSo rearrange to solve for time: (v-u)/a = t = 14.98 seconds.\n\nHope that helps. And learn your equations!
912139
The abverage acceleration is a = (V^2)/2x = 35^2/(2*.11) = 5568.18 m/s^2\nMultiply this by the mass and you get the average force:\n\nFa = .14*5568.18 = 779.14 N
248389
The speed of sound increase at a rate of 0.6-m/s per degree C starting at 0 degrees C with a speed of 330-m/s.  At 30 degrees C the speed of sound will be 330-m/s + (0.6-m/s*30)=348=m/s
33077
What is the cop car's acceleration? Do they both start from standing?\n\nEDIT\n5 seconds\n\nSay the car is at position 0m, and the cop is at -90m\n\nAfter 5 seconds, the cop has done 20*5=100m, so is at position 10m\n\nThe car has accelerated to 4m/s, giving it an average speed over this time of 2m/s. This avg speed over 5 seconds means he has travelled 10m, putting him also at position 10m, alongside the cop.
1044764
The law of conservation of momentum holds in this case.\n\nThe initial momentum of the system must equal the final momentum of the system.\nIn this case, we will call the "System" both cars.\n\nCar 1 has a mass (m1) of 1000 kg, velocity (v1) of 20 m/s.\nCar 2 has a mass (m2) of 800 kg and a velocity, in what we will call the negative direction, v2 = -30 m/s.\n\nThe total momentum (P) does not change.\n\nP = P1 + P2\n\nP1 = m1 * v1 = 1000 kg * 20 m/s\nP1 = 20000 kg m/s\n\nP2 = m2 * v2 = 800 kg * -30 m/s\nP2 = -24000 kg m/s\n\nP = 20000 kg m/s + -24000 kg m/s\nP = -4000 kg m/s\n\nSo the resulting 1800 kg mass will have a momentum of -4000 kg m/s, meaning that it will be traveling in the "negative" direction, AKA, South.\n\nP = -4000 kg m/s = 1800 kg * v_f\nv_f = -4000 / 1800\nv_f = -2.22 m/s\n\nThe resulting blob of the two cars which stuck together will be traveling -2.22 m/s South.
728641
acceleration can be found using\na=(m1-m2)g/(m1+m2) = 3*9.8/5 = 5.88 m/s^2
918000
Ok, iam assuming you know the equations of motion: \nv= u + at\ns=ut + 0.5at^2\nv^2=u^2 + 2as\ns=((u+v)/2)t\n\nWe need to break this motion up into seperate sections.\n\nNote however that all the units in this question are in SI units, so they are consistent with each other. You don't have kilometres which you would need to convert into metres for example, before using these equations. In that case I won't put in the units by the numbers in my calculations everytime, just to make it look clearer. But they are there and they are consistent.\n\n\nFirst section is from rest (u=0) accelerating at 2.6m/s^2 for 18s. We will find the displacement travelled here.\n\nWe know u(initial velocity)=0, we know a(acceleration)= 2.6 and t=18 and we want s(displacement)\n\nthe equation connecting these four is s=ut + 0.5at^2 \n(we don't need v and this equation does not have v)\n\nso s = (0 x 18)  +  (0.5 x 2.6 x 18^2)\n       = 421.2m << note this displacement, we're going to use it at the end\n\n\nThe next section, we have the automobile continuing its motion at a constant velocity that it has gained after accelerating at 2.6m/s^2 for 18s.\n\nHere, we need to find what this constant velocity is, which is easy. \n\nSince the body is accelerating at 2.6 metres per second every second it is gaining a velocity of 2.6 metres per second, every second.\n\nSo the velocity it has gained after it has accelerated for 18s is just \n2.6m/s^2 multiplied by 18s \n= 46.8m/s^1\n\nIt's important you notice that I am using 18s because this is how much time it has accelerated for to reach that constant velocity; it stopped accelerating after 18s. 15s is the time it travels at that speed it has gained after accelerating for 18s.\n\nNow this velocity is constant so every second it travels a displacement of 46.8m, the displacement it travels here is just 46.8m/s^1 multiplied by 15s\n= 702m << note this as well.\n\n\nWe have one last section now. Here, it is accelerating at -2.9m/s^2 from a velocity of 46.8m/s^1 (the constant velocity we calculated remember) to a velocity of 0m/s^1 (rest). So we need one of the equations of motion here.\n\nWe know u=46.8m/s^1, a= -2.9m/s^2 and v must = 0m/s^2 \nwe want s. No t is involved here so the equation that doesn't have t is \nv^2=u^2 + 2as\n\nThis equation isn't giving us displacement in this form though. We need to rearrange to make s the subject by leaving it alone on one side of the equal sign.\n\nFirst lets put u^2 to the other side of the equal sign (iam taking it away from both sides of the equal sign to do this). \n\nThat gives us v^2 - u^2=2as\nNow we just need to divide both sides by 2a to leave s on its own.\nThat gives us \n\n(v^2 - u^2)/2a = s\n\nNow referring back to our values, we just subsitute them in,\nwhich gives us\n\n(0^2 - 46.8^2) / (2 x -2.9) = 377.6275862m << note this lastly\n\nRemember all them displacements that we noted, now all we have to do now is add them together. To find the total displacement travelled. That is \n\n421.2m + 702m + 377.6275862m = 1500.827586m\nWhich is your answer\n\nA few notes here:\nThe equations we used gives us displacement only, not distance travelled. That's why I always referred to them as displacements. And same for velocity. The equations only deal with velocity, not speed. The difference is that speed doesn't care about it's direction, velocity does. However, the automobile just travels in a straight line and straight forward in this case; there isn't any case where it travels backwards or anything. It just travels in one direction, so displacement automatically gives us distance travelled anyway and velocities in this case is the same for speed.\nAlso I didn't round up the any numbers (especially that long 377.62... number) becuase I don't want to round numbers until I get to the final answer, becuase that will introduce inaccuracies. I can however round up now at the final answer. \nI leave tha
859010
The average veocity of the car from 0 to 25 m/s is 12.5 m/s.\n\nThe total amount of time it takes is 200 m / (12.5 m/s) = 16 seconds.\n\nThe change is speed in 16 seconds is from 0 to 25 m/s.  Therefore, the acceleration is (25 m/s)/ (16 s) = 1.5625 m/s^2\n\nThe force = mass * acceleration = 1600 kg * 1.5625 m/s^2 = 2500 N\n\nAnswer is d.
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...
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
979348
acc is change in velocity divided by time taken.\nso the change in velocity is 17m/s and time taken is 4 sec.\ntherefore 17/4 =4.25m/s2.\nthus the ans is C.