A jeepney from rest accelerates uniformly over a time of 7.25 s and covers a distance of 15 m. Determine the acceleration of the jeepney?
Answer (1)
[tex]\textbf{Given:} \\[/tex][tex]v_0 = 0 \text{ m/s} \\[/tex][tex]t = 7.25 \text{ s} \\[/tex][tex]d = 15 \text{ m} \\[/tex][tex]\textbf{Formula:} \\[/tex][tex]d = v_0t + \frac{1}{2}at^2 \\[/tex][tex]\textbf{Substitute and solve for } a: \\[/tex][tex]15 = (0)(7.25) + \frac{1}{2}a(7.25)^2 \\[/tex][tex]15 = \frac{1}{2}a(52.5625) \\[/tex][tex]15 = 26.28125a \\[/tex][tex]a = \frac{15}{26.28125} \\[/tex][tex]a \approx 0.57 \text{ m/s}^2 \\[/tex][tex]\textbf{Answer:} \\[/tex][tex]\text{The acceleration is approximately } 0.57 \text{ m/s}^2.[/tex]
