Cheetahs, the fastest of the great cats, can reach 45 {\ m mph} in 2.0 {\ m s} s

Question

Cheetahs, the fastest of the great cats, can reach 45 { m mph} in 2.0 { m s} starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration (in (a) m/s^2 and (b) ft/s^2) and the distance (in (c) m and (d) ft) they travel during that time.

Explanation / Answer

vf = 45mph = 45*1609m/3600s = 20.11 m/s a) let acceleration be a, we have: vf = at => a = vf / t = 20.11 m/s / 2.0s = 10.06m/s2 b) a = 10.06 m/s = 10.06*3.28 ft/s2 = 32.98ft/s2 c) distance D = (1/2)at2 = (1/2)*10.06m/s2(2.0s)2 = 20.12 m d) Distance D = (1/2)at2 =0.5*32.98ft/s2 (2.0s)2 = 65.96 ft PLEASE RATE.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.