a car travels at 55 miles per hour. Is it fundamental or derived quantity?
397194
it is a derived quantity obtained from the fundamental units of length and time.
811400
If you ignore wind resistance the calculation is quite straightforward.\n\nAcceleration due to gravity is 9.8 metres per second per second.\nEvery second the car will accelerate by an additional 9.8 m/s\n\nYou convert 100 km/hr into m/s first:\n\n100 x 1000 (metres in a Km) / 3600 (seconds in an hour) = 27.78 m/s\n\nDivide that speed by the acceleration:\n\n27.78 / 9.8 = 2.83 seconds
718292
40 miles per hour.  For instance if it is a 120 mile trip (one way) it will take 4 hours there, and 2 hours back.  \n\n6 hours for a 240 mile round trip.  6/240 is 40 mph.
685470
the speed = distance/time
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.
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.
873979
deceleration = change in speed / time\ndeceleration = (50 (km / hr)) / (0.15 s) = 92.5925926 m / s2\n\nF = ma\nF = (1 500 kg) * 92.5925926 (m / (s^2)) = 138 888.889 newtons
306765
Speed is always relative to something else.\n\nSo, relative to the car, the fly is going at quite a low speed as it buzzes around inside.\n\nRelative to the road, the fly is going at just over or just under 70mph, depending on which direction it is flying within the car.\n\nRelative to the sun, the fly is moving at thousands of miles per hour as it rotates once around the sun every year, just like the rest of us.
716944
If there is no net force (i.e. sum of forces is zero) then there will be no change in velocity. So if the object was going 15 m/s and the net force is zero, then it will still be going 15 m/, (for 2 sec = 30 meters traveled). Same concept for your jet car - constant velocity means no net force ( there may be forces acting on it, but if the velocity is constant, they must sum to zero - net force is zero).\n\nAs Newton's first law says, "An object in motion tends to stay in motion unless acted on by an outside force." So constant velocity (i.e. stays in motion at the same velocity) means no net force acting on the object.
403577
You have miles per minutes, and you want to convert it to miles per hour.  Set it up as fractions so the units cancel out, like this:\n\n1mile/8minutes * 60minutes/1hour  Now the minutes cancel and miles/hour is the final units.  Now just do the math.  60/8 = 7.5 mph.
704006
assuming no one else jumps out of the car (at over 100mph!!!) then you work it out in two parts like this:\n\ntime = distance/velocity\n\ni) t1 = 10miles/105mph = 0.095hours\n\nii) t2 = 40miles/130mph = 0.307hours\n\nadd t1 and t2 for total time = 0.095 + 0.307 = 0.4hours = 24 minutes. (60mins/10 and then *4)
232197
24,900 miles / 24hrs = 1037.5mph\n40,076 km / 24hrs = 1669.8kph = 463.83m/s
327132
Its airspeed will remain the same.\n\nIts groundspeed will be the vector sum of its airspeed and the car's speed.
160887
The insect is inside the bus, and so is moving 5 mph relative to the bus (that is what you meant right?) so:\n\na. 5 mph\nb. 55 mph, since the insect is moving 5 mph faster than the bus\nc. makes no sense since they are moving relative to each other
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.
749570
V = √(550² + 80²)\n≈ 556 mph\nDirection tanθ = 550/80 in 3rd quadrant\nθ ≈  81° 43' 27" from West\n= 188° 16' 33" T (clockwise from N)
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