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2.) From rest a jeepney accelerates uniformly over a time of 3.25 s and covers a distance of 15 m. Determine the acceleration of the jeepney.

Formula: speed = distance x time; a =  \frac{ v_{f}-  v_{i}  }{t}


s = dt
   = (15 m)(3.25 s)
   = 48.75

a =  \frac{ v_{f}- v_{i} }{t}
   =  \frac{48.75 m/s - 0}{3.25 s}
   =  \frac{48.75 m/s}{3.25 s}
   = 15 m/s²

The acceleration of the jeepney is 15 m/s².

3.) A train accelerates a speed of 20 m/s over a distance of 150 m. Determine the acceleration of the jeepney.

Formula : time =  \frac{distance}{speed} ; a =  \frac{ v_{f} -  v_{i}  }{t}


t =  \frac{d}{s}
  =  \frac{150 m}{20 m/s}
  = 7.5 s

a =  \frac{ v_{f} - v_{i}  }{t}
   =  \frac{20 m/s - 0}{7.5 s}
   =  \frac{20 m/s}{7.5 s}
   = 2.667

The acceleration of the train is 3 m/s².

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