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Question:I need it for a quiz all at once. Thank you.

Answers:I keep this cheat sheet handy for quick reference when answering at YA. It refers to constant acceleration with initial velocity = 0. x = distance, a = acceleration (e.g., g) x, v, a and t are each given in terms of the possible combinations of 2 of the other 3 variables; e.g., x is given as f(a,t), f(v,t) and f(v,a). x = at^2/2 = vt/2 = v^2/(2a) v = at = 2x/t = sqrt(2ax) a = v/t = 2x/t^2 = v^2/(2x) t = sqrt(2x/a) = v/a = 2x/v Note that the above will work just as well for rotation. Substitute theta for x, omega for v and alpha for a.

Question:I understand the basic equation i.e Rate = Distance over Speed Distance = Rate x Speed Speed = Distance over Time But when it comes to Word Problems like these 3 2 vehicles moving towards each other, when will they meet or whats their speed? 2 vehicles moving away from each other 2 vehicles going in same direction, one started before the other, how far are they from the start Does anyone have any good ways of remembering the formulas for these? There appears to be addition and subtraction involved.. any tips on remembering or figuring out these?

Answers:1. if the vehicles are moving in the same direction, subtract to get the relative speed 2. if the vehicles are moving in opposite directions [either towards each other or away from each other] add to get the relative speed

Question:It was once recorded that a Jaguar left skid marks which were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2 (superscript 2), determine the speed of the Jaguar before it began to skid. I'm really stumped on this question! How would I calculate speed when I am just given a distance and the acceleration? Especially when they want to know how fast the Jaguar was going before it started to skid.

Answers:The easiest formula to use, one that relates distance, velocity and acceleration without using time, is this one: v^2 = v0^2 + 2*a*d where v0 = initial velocity, v = final velocity (zero in this case), a = acceleration, d = distance. But you can always use time to get there also. Solve d = (1/2)at^2 as if it were accelerating rather than decelerating over this distance. The time would be the same to accelerate to v0 as to decelerate from v0. Now, once you know that time, plug it into v = v0 + at You'll get the same answer.

Question:I need to know the formula in physics when you're looking for time, and you're given the acceleration and the distance.

Answers:v = u + at v2 = u2 + 2as s = ut + 1/2 at2 where s is the distance, u is a speed , t is time and a is acceleration, plug in your values and calculate :-p average velocity = (v + u)/2

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Rate-Time-Distance Problem 3 :Videos by Julie Harland yourmathgal.com Solves this word problem using uniform motion rt=d formula A 555-mile, 5-hour trip on the Autobahn was driven at two speeds. The average speed of the car was 105 mph on the first part of the trip, and the average speed was 115 mph for the second part. How long did the car drive at each speed? Answer: 105 mph for 2 hours and 115 mph for 3 hours