I’m pretty sure the answer is test
We will be using this equation for this problem
d = ut + ½.at²
<span>Given:</span>
<span>initial velocity, u = 0 (falling from rest) </span>
<span>acceleration, a = +9.80 m/s²(taking down as the convenient positive direction) </span>
<span>Time = 1.0s, 2.0s, 3.0s, 4.0s, 5.0s </span>
<span>Using .. d = ½.at² each time (each calculation is the distance from the top) </span>
<span>For 1.0s .. d = ½ x 9.80 x (1²) = 4.90 m </span>
<span>For 2.0s .. d = ½ x 9.80 x (2²) = 19.60 m </span>
<span>3.0s .. d = 44.10m (you show the working for the rest) </span>
<span>4.0s .. d = 78.40 m </span>
<span>5.0s .. d = 122.50m </span>
<span>Plot distance (displacement from the top) on the y-axis against time on the x-axis (label axes and give units for each).The line of best fit will be a smoothly upward curving line getting progressively steeper. Do not join graph points with straight lines.</span>
Ok so, based on the graph lets say that x = seconds and y = depth of dolphin. the interception in both points is when they are at 0 i.e when x is equals 0 and y equals 0. So At 0 seconds, the dolphin is 42 feet below the surface. So we say that x = 0, y = -42 and then the y intercepts = -42 so the point of interception is(0 (seconds),-42(depth of the dolphin)) When the clock says it's 14 seconds, the dolphin is even with the surface (0 below the surface this time). So we say that x = 14, y = 0. In this case The x intercept = 14 (14,0). But we need to calculate The slope with the formula= (y2 - y1)/(x2 - x1) = 42/14 = 3. Therefore, the formula for this line is y = 3x - 42.
I dont understand, is it 15% over 15 % or just 15 % or is it 293$ over 293$ or is it just 239.
