4) a racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) if it’s moving at 53 m/s at the end of this interval, what was its speed at the beginning of the interval? (b) how far did it travel from rest to the end of the 140-m distance

Answer

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

       v₀ + 53 = 77.78

        v₀ = 24.78 m/s or 25 m/s

b)

         a = 7.8 m/s²

using equation of motion

  v₀² = v₁² + 2 a s

  53² = 0²+ 2 x 7.8 x s

  s = 180 m

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example