speed and acceleration?
931347
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.
193077
Force=Mass x Acceleration\n\nSo Acceleration = Force/Mass\n\nHope this helps
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
736848
Velocity is basically speed, normally a constant speed used when measuring extremely fast objects such as bullets. Acceleration is the rate at which an object increases speed. For example a car can accelerate from a standstill to 60mph in ten seconds
919955
Speed and Acceleration are related in that acceleration is the rate of change of your speed (should be velocity since you can take into account change in direction for accelerations).\n\nThe relationship can be mathematically expressed as:\ndv/dt = a.\n\nThat is, the change in velocity over the change in time.
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
475727
the answer to the first question is 5 m/s.\nF=ma; a=F/m=200N/40kg=5m/s\nthe second answer is 1m/s.\nF=ma; a=F/m=2N/2kg=1m/s\nthe third answer is 1:4.\nF=ma; m=F/a\n(1) M=f/a=F/a=m\n(2) M=f/a=4F/a=4m
826560
3.0m/s is not an acceleration, it is a speed.\nIs the acceleration constant?
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
963172
average speed =total distance/total time\n1.\ntotal distance=2km\ntotal time=1/2hr\naverage speed =4km/hr\n2.\ntotal distance=360km\ntotal time=5hr\naverage speed =360/5=72km/hr\nthanks & good luck
999483
Multiply by time...\n\nMeters per second squared is an acceleration... Which is velocity/time.\n\nMultiplying by time will get you velocity... Which is in meters/second.
978564
Velocity is the speed an object is moving at and the direction that it is moving in. For example, if you said that a car was going 60 mph due east you would be talking about velocity; if you merely said that the car was moving 60 mph you would just be talking about its speed. You would measure acceleration by measuring an objects speed at two different times (the closer the more accurate your number estimate), subtract the difference and divide by the time interval. For example, if at time=1, the speed of an object was 5 meters per second, and at time=4 the speed was 14 meters per second you would do the following: (14-5)/(4-1)=9/3=3 meters/sec^2.\nRegarding  your last question, if the skateboard was going 5mph when you started your experiment, then your initial speed would b 5mph.
65071
Force = mass x acceleration\n Mass = weight times gravity\nAcceleration is speed/time\n example if I’m playing football and I want to move the other player assuming we are both on the same planet (gravity is constant) and we are both moving at the same speed at each other I can do so if am heavier than him (increase wt (g) = increase mass) or I decease the time it takes to reach that player (constant speed and distance in less time = acceleration) or I can do both. This is how you increase force by increasing acceleration, mass, or both
441442
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
823160
first problem: let current=c and the speed of the kayak=k\n\nk+c=4.7\nk-c=3.4        add the two equations.\nso, 2k=8.1    divide by two.\nso, k=4.05    plug this into one of the equations.\n4.05+c=4.7   subtract by 4.05\nc=.65          \n \nboth answers in mph.\n\nuse this same logic for problem two.
185601
Another way to see it is to consider acceleration versus constant speed. If you're going 60mph your position is increasing at that constant rate, but if you're pushing on the accelerator your speed is going 60,61,62 etc.. so your speed (which is your position vs time rate) is increasing. That is a varying rate. Doesn't have to be strictly increasing though. You could be speeding up then slowing down then back up again. Or maybe accelerating or decelerating faster, then slower. It can be any combination as long as you're not constantly going at one speed.
750537
acceleration = v^2/r or v*v/r (this is m/s^2 so divide by 9.81 to get g's)\n\nwhere v is the velocity\n\nr is the radius\n\nBe careful, think about the one g the pilot feels from the pull of the earth. ;)