5)What is the total time the hare is moving? 6)What is the acceleration of the t

Question

5)What is the total time the hare is moving?

6)What is the acceleration of the tortoise?

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 0.9 m/s2 for 4.7 seconds. It then continues at a constant speed for 11.6 seconds, before getting tired and slowing down with constant acceleration coming to rest 71 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. uniformly for the entire distance, finally catching the hare just as the hare comes to a stop.

Explanation / Answer

for the first part

time = t1 = 4.7 s

speed of the hare after 4.7s = v

v = vo + a1*t = 0 + 0.9*4.7 = 4.23 m/s

distance travelled by hare in 4.7 s = x1 = vo*t + 0.5*a*t^2 = 0 + (0.5*0.9*4.7^2)

x1 = 9.94 m

for the second trip

distance travelled by hare in 11.6s = v*11.6 = 49.1 m

time t2 = 11.6 s

for the third trip

final speed = vf = 0

initial speed vi = v = 4.23 m/s

distance travelled before coming to rest = x3 = 71 m

from

vf^2 - vi^2 = 2*a*x

0^2 - v^2 = 2*a2*x3

a2 = -0.126 m/s^2

t3 = v/a2 = 33.6 s

4) a = -0.126 m/s^2    <<<----answer

5) T = t1 + t2 + t3 = 4.7+11.6+33.6 = 49.9 s    <<<----answer

6)

total distance x = x1 + x2 + x3 = 130.04 m

x = vo*T + 0.5*a*T^2

130.04 = 0 + (0.5*a*49.9^2)

a = 0.104 m/s^2   <<<----answer

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