22. a - (vf^2 - vi^2)/(2d) a = (0 - 23^2)/(170) a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2 33-0/t = 3 33 = 3t t = 11 sec to reach 33 m/s^2 Find the av velocuty: 33+0/2 = 16.5 m/s Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from: (Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2 The elapsed time is found by solving Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2 S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2 final velocity, v = 0m/s acceleration, a = -7.00m/s^2 displacement, s - 92m Using v^2 = u^2 - 2as 0^2 - u^2 + 2 (-7.00) (92) initial velocity, u = sqrt (1288) = 35.9 m/s This is the speed pf the car just bore braking.
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