time taken for changing velocity is 5min.\nwhat is acceleration?.(there will be acceleration because velocity is changing)
131355
We have to find the resultant acceleration of acceleration in East direction and acceleration in the North direction, since there is change in direction.\n\nSince it moves with same speed in both directions it is enough if we find in any one direction.\n\nFor the first 2.5 s there is no acceleration, similarly there is no acceleration in the second 2.5 s. \n\nAt the end of first 2.5 s it has to come to stop and then turn to north all with zero time. \n\n In the Eastern direction the change in speed is 5- 0 = 5m/s. To find acceleration divide this by zero.  It is not defined. \n\nIf we assume that it takes a very small time, then the acceleration in this direction is nearly   minus infinity toward west (as the speed is reduced).\n\nSimilarly in the north direction it is plus infinity toward west.\n\nThe resultant of these two is inclined toward North West with a magnitude of infinity.
504239
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
154830
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
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.
482098
f=ma\n\na=change in velocity/change in time = (5.53-2.25)/1 \n\nm= 57\n\nf = 57 (5.53-2.25)\n\npunch the numbers in to your calculator, and there it is!
463496
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2
388140
I'm not exactly sure what 'time frame' this question is asking for (that is if it's asking for the entire reentry process).  I'll show you how to get the two intervals' acceleration, and if it's asking for the accel of the entire process, you can use my method to find the answer. ^_^\n\nMain equations here:  \na (acceleration) = change in velocity / change in time\nchange in velocity = final velocity - initial velocity\nchange in time = final velocity - initial velocity\n\n\n~~~~\n\nThe first acceleration is during the initial reentry.\n\nThe equation for finding acceleration requires finding the change in velocity and also the change in time.  So do that first.\n\nchange in velocity = final velocity - initial velocity\nchange in velocity = (496km/hr) - (565km/hr)\nchange in velocity = - 69 km/hr\n\nchange in time = final time - initial time\nchange in time = (15 seconds) - (0 seconds)\nchange in time = 15 seconds\n15 seconds = 0.25 hours\n\na = change in velocity / change in time\na = (-69 km/hr)/(0.25 hr)\na = - 276 km/hr2 (kilometers per hour squared)\n\n~~~~\n\nThe second interval:\n\nchange in velocity = final - initial\nchange in velocity = (346km/hr) - (496km/hr)\nchange in velocity = - 150 km/hr\n\nchange in time = final - initial\nchange in time = (17.0 seconds) - (0 seconds)\nchange in time = 17.0 seconds\n17.0 seconds = 0.283 hr\n\na = change in velocity / change in time\na = (-150 km/hr) / (0.283 hr)\na = -530 km/hr2 (kilometer per hour squared)
624366
1)Acceleration is the change in velocity (speed) per unit time\n\nVelocity = (meters/second)\nAcceleration = \n(meters/second) / second = meters/second-squared\n\nAcceleration = meters per seconds squared or D\n\n2)B\n\nFinal position = \n(initial position) + (initial velocity * time) + (-9.8/2 * time squared) / 2\n\nX is the height of the ledge\nInitial velocity is zero and so is final position:\n\n0[m] =  X[m]  + 0[m/s] - (4.9[m/sec sqrd] * 64[sec sqrd)\n\nX = 313.6 [m] or meters
470903
velocity is given by :\nv(t) = -1 + e^(1-t)\n\nacceleration is given by :\na(t) = - e^(1-t)\n\nSo :\na) \nv(3) = e^(-2) -1\na(3) = - e^(-2)\n\nb) at time t = 3, acceleration is negative. The speed of the particle is decreasing.\n\nc) whenever the particle changes direction, it is velocity is momentarily zero.\n\n=> 0 = -1 + e^(1-t)\n=> 1 - t = 0\n=> t = 1\n\nNote that velocity is positive for all times < 1, and is constantly decreasing. For all times > 1, velocity is negative. Thus, the particle changes direction at time t = 1.\n\nd) at time t = 0, \nx(t) = - e\n\nat time t = 1,\nx(t) = -1 - 1 = -2\n\ntotal distance travelled thus far = (e - 2)\n\nat time t = 3,\nx(t) = - 3 - e^(-2)\n\ntotal distance by the particle over the interval 1<=t<=3 is :\n1 + e^(-2)\n \ntotal distance travelled is :\n(e - 2) + (1 + e^(-2))\n= e - 1 + e^(-2)
226476
This is more of a physics question...\nd final = .5*a*t^2 + v*t + d\nv final = a*t + v\nd is displacement, v is velocity, a is acceleration.\nEnjoy.
219418
So long as there is rate of change of motion
374334
Velocity(at time t) = Velocity (at time = 0) + Acceleration x Time\n\nV(t) = Vi + At\nV(t) = 32 ft/sec + (-32 ft/sec/sec)(t)\nV(1 sec) = 32 - 32 = 0 ft/sec\n\ny(t) = y(initial) + (V x t) + 1/2(A)(t^2)\ny(1 sec) = 0 + (32) +  (-16) = 16 feet\n\nTo hit ground, y = 0, so (V x t) = 1/2(A)(t^2)\n32t = 16t^2\n2 = t, t = 2 seconds\n\ny(max) occurs at zero velocity, which is at t = 1 sec, x = 16 feet
406121
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! :)
436232
3.0m/s is not an acceleration, it is a speed.\nIs the acceleration constant?
433053
wish I could solve em for you but ill try and help still.\n\n1) The acceleration should be the derivative of the equation for the car's velocity.  The original distance/time equation would be the anti-derivative of the velocity.  \n\n2 The equation to find out how long it will take to get there is h+ax-16x^2 in feet.  (a is the original speed and x is number of seconds and h can be the height of the building.)\n\nHope this helps
105407
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
549913
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.