An 899-kg (1987 lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s. (a) find the average acceleration of the dragster during this time interval. 44.58 correct: your answer is correct. m/s2 (b) what is the magnitude of the average net force on the dragster during this time

Explanation:

Mass, m = 899 kg

Initial velocity, u = 0 m/s

Final velocity, v = 26.3 m/s

Time, t = 0.59 s

a) We have equation of motion v = u + at

   Substituting

                     v = u + at

                     26.3 = 0 + a x 0.59

                       a = 44.58 m/s²

     Acceleration of dragster = 44.58 m/s²

b) Force = Mass x Acceleration

   F = ma

   F = 899 x 44.58 = 40074.07 N

   F = 40.07 kN

  Average net force on the dragster during this time is 40.07 kN

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.