Consider the following velocity versus time graph for a moving car
velocity (m/s)
-304
0
10
20
30
40 50 60
70
80
what would the car's position be at 80.0 s if it started at a position xo = -10.0 m at t = 0 s?
(a)
- 560 m
(b)
-550 m
(c)
960 m
(d)
950 m

Simple harmonic motion wind energy circulation best answer from the choices

It would stay the same

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.