A tortoise and hare start from rest and have a race. As the race begin both acce

Question

A tortoise and hare start from rest and have a race. As the race begin both accelerate forward. The hare accelerates uniformly at a rate of 0.8 m/s^2 for 4.3 seconds. It continues at a constant speed for 13.1 seconds, before getting tired and slowing down with constant acceleration eventually coming to rest 56 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop. How fast is the hare going 1.7 seconds after it starts? How fast is the hare going 12.4 seconds after it starts? How far does the hare travel before it begins to slow down? What is the acceleration of the hare once it begins to slow down? What is the total time the hare is moving? What is the acceleration of the tortoise?

Explanation / Answer

a) Accleration till 1.7s a = 0.8 m/s2, intial velocity vi = 0, vf = ?, t = 1.7s

using v = u + at => v = 0 + 0.8*1.7 = 1.36 m/s (speed of hare at 1.7 s)

b) To measure speed at 12.4 s we first calculate speed at 4.3 s (v1) (where accleration is 0.8 m/s2) and then use it to calculate speed at 12.4s (vf)

v1 = u + at => v1 = 0 + 0.8*4.3 = 3.44 m/s, Since after this hare moves with constant speed hence the speed of hare at t = 12.4s is 3.44 m/s.

c) Let s1 be distance travelled by hare in 4.3 s and s2 be the distance travlled in 13.1 s

s1 = ut + 1/2*a*t2 = 0*t + 0.5*0.8*(4.3)*4.3 = 7.4 m

and s2 = ut + 1/2at2 = 3.44*13.1 + 1/2*0*4.32 = 45.1 m

Hence distance before hare starts to slow down is s1 + s2 = 7.4 + 45.1 = 52.5 m

d) Hare comes from 3.44 m/s to zero velocity in 56 - 52.5 = 3.5 m

Hence using v2 = u2 + 2aS => 0 = 3.442 + 2*a*3.5 => a = -1.7 m/s2

e) Time required to slow comes to halt after 52.5 m

v = u + a*t => 0 = 3.44 - 1.7*t => t = 2 s

Hence total time taken by hare = 4.3 + 13.1 + 2 = 19.4 s

f) If accleration of tortoise is a then it took tortoise 19.4 s to travel a distance of 56 m

Hence using S = u*t + 1/2*a*t2 => 56 = 0*19.4 + 1/2*a*(19.4)2  => a = 0.297 m/s2 = 0.3 m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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