He decides to test the car by racing with another speedster, john. both start from rest,
but experienced john leaves the starting line 1.00 s before peter. if john moves with a
constant acceleration of 3.50 m/s2, and peter maintains an acceleration of 4.90 m/s2,
find (a) the time it takes peter to overtake john, (b) the distance peter travels before he
catches john, and (c) the speeds of both cars at the instant peter overtakes john.

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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.