From Rest, A Jeepney Accelerates Uniformly Over A Time Of 3.25 Seconds And Covers A Distance Of 15 Cm

From rest, a jeepney accelerates uniformly over a time of 3.25 seconds and covers a distance of 15 cm

 

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Problem:

From rest, a jeepney accelerates uniformly over a time of 3.25 seconds and covers a distance of 15 cm

Given Data:

Vi (initial velocity) = 0 m/s

d (distance)  =  15 cm or 0.15 m  (15 cm × 1 m/ 100 cm =  0.15 m)

t (time) = 3.25 s

a (acceleration) = ? unknown

Solution:

We can use the formula a = d ÷ 1/2(t)² since Vi = 0 m/s

a = d ÷ 1/2(t)²

  = 0.15 m ÷ 1/2 (3.25s)²

  = 0.15 m ÷ 1/2 (10.56 s²)

  =  0.15 m  ÷ 5.28 s²

 = 0.028 m/s²

Answer:

acceleration is 0.028 m/s²