Question

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 1

Answer

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

    $D = \dfrac{v_0 + v_1}{2}\times t$

    $140 = \dfrac{v_0 + 53}{2}\times 3.6$

       v₀ + 53 = 77.78

        v₀ = 24.78 m/s or 25 m/s

b) $a = \dfrac{v-u}{t}$

   $a = \dfrac{53-25}{3.6}$

         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