distance speed time formula triangle
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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.
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
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.
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