What is the total distance travelled for an accelerating automobile?
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
893501
44.44444 mph.  If the distance of one leg is x miles, then you drove a total of 2x miles in (x/40 + x/50) hours.  Divide the total distance driven by the total time spent driving, and you get 200/9, or 44.444... mph
607556
Everyone is wrong, it is 48MPH, think of it this way.  Lets say the distance is 60 miles, it takes 1.5 hours at 40, and 1 hour at 60, so a total of 2.5 hours, divide 120 by 2.5 and it is 48mph.  This will always be the answer reguardless of distance, whether 1 mile in each direction or a million miles in each direction.
202957
How far apart are the marks?\nTime taken to travel between the marks?\nSpeed = distance divided by time.
798019
Since car 1 drove for 2.5 hours at 90 km/h, car 1 went 225 km.  So it's clear that L is not on the line between A and B, and that the distance from A to L is 225 km.  The distance from A to B is 80 km.  There must be something in the picture that tells you one additional piece of information about the triangle ALB, that will allow you to calculate the distance from B to L.  That distance, divided by 2.5 hours, is the speed of car 2.
94585
I checked with AAA.com and the driving distance is 2861 mi (4605 km) and would take about 41 hours. If you have a AAA membership you can make a Triptik travel planner on line with maps, rest stops, hotels, restaurants and gas stations all noted.
10487
1,000,000,000,000km\n621,371,192,237miles
782291
Assuming that the friction with the ground provides a constant force, the clumped-together vehicles decelerate at constant rate until they stop. Because we assume the ground to be horizontal, that rate is (mu) * g, where (mu) is the coefficient of friction. Therefore the rate of deceleration is .923 g or 9.052 m/s^2.\n\nThe distance traveled is 18 m. For constant acceleration (or deceleration, it does not matter as long as the velocity at one end of the time period is zero) x = 1/2 a t^2. We know x and a; therefore we find t = 1.994 sec.\n\nNow we use this time and the assumption of constant acceleration to find that the initial velocity was v = a * t = 18.05 m/s. Therefore the intial momentum was 18.06 * 3000 = 54 150 kg m/s, directed toward the northwest.\n\nTo cause that momentum, the momenta of the individual vehicles added, and they are vectors. The truck was headed north, the car, west, and they ended up going northwest. Therefore they had equal-magnitude momenta. (The tangent of the resultant angle is the ratio of the two.)\n\nSay Pt = truck momentum = Vt Mt, Pc = Vc Mc.\nDealing just in magnitudes for the moment, \nPt = Pc --> Vt Mt = Vc Mc, so Vc = Vt (Mt / Mc) = 2.75 Vc.\nBecause the vehicles started at right angles to each other, the magnitude of the final momentum is Pf = sqrt(Pt^2 + Pc^2) = \nsqrt((Mt * Vt)^2 + (Mc * 2.75 * Vt)^2) = 3111 * Vt\nWe have already found Vf = 18.05, and we know the combined mass is 3000 kg; therefore we infer Vt = 18.05 * 3000 / 3111 = 17.41 m/s. Therefore Vc = 47.87 m/s.
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
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
990022
the question is not so clear. as per my understanding it takes 7 sec to go from 0-60miles/hour. how many seconds it takes to go from 0 to100 miles/hour?\n\n(100 multiplied by 7) divided by 60.\n\ni.e. 100*7/60=11.6 sec approx.
30694
20m/s.\nbecause the total time is 5s. and by defn of average speed it is the ratio of total distance travelled to total time taken i.e 100/5 = 20m/s
157405
Ion drive engines can reach speeds of about 30,000 miles per hour.  If someone lived 80 years travelling at that speed they would travel 21,024,000,000 (about 226 times the distance to the Sun from Earth).
393461
First put everything in the same units.\n\n36 km/h * 1000m/1km * 1hr/3600s = 10m/s\n\nOk...the equation for on objects position as a function of time is.\n\nx(t) = xo + v*t + a*t^2\n\nwhere \nxo = initial position (0 for both car and truck)\nv  = velocity (10 m/s for the truck)\na  = acceleration (4 m/s2 for the car)\nt  = time\nx(t) = position at any given time t\n\nIn question 1 you are looking for the time at which the car and the truck are at the same position.  Or the time at which x(t) for the car equals x(t) for the truck.  So set the position equations equal to each other and solve for t.\n\ncar equation:  x(t)car = 0 + 0 * t + 4 * t^2\ntruck equation: x(t)truck = 0 + 10 * t + 0 * t^2\n\ntherefore when they are at the same position\n\n4t^2 = 10t\n\nor\n\nt = 10/4s or 2.5s (note: the answer you have given is incorrect.  At 5 seconds the car would be 100 meters from the light and the truck would be 50 meters from the light)\n\nFor question 2 we need to take the derivative or the position equation for the car and solve for the velocity at time t = 2.5\n\nx(t)car = 4t^2\n\nd(x(t)car)/dt = v(t)car = 8t\n\nv(t)car (at t = 2.5) = 8*2.5 = 20 m/s  your answer for this question proves to be correct
295799
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
741279
Total distance traveled divided by total time in transit.\n\nr = d/t.
237365
Say that x is 10 miles.\n\nIt takes 10/10 hours on the first trip (1 hour)\nIt takes 10/8 hours on the return trip (1.25 hours)\n\nHe has therefore travelled 20 miles in 2.25 hours.\n\n20/2.25 = 8.89 miles per hour.