Question

4) 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 1.1 m/s² for 4.1 seconds. It then continues at a constant speed for 13.7 seconds, before getting tired and slowing down with constant acceleration coming to rest 81 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.

a) What is the acceleration of the hare once it begins to slow down?

b) what is the total time the hare is moving?

c) what is the acceleration of the tortoise?

Answer 1

Solution:

During first 4.1 s

Speed, v=vo+a*4.1

Replacing with data:

v = 0 +1.1*4.1 = 4.51 m/s

Now , we have to find the position at time 17.8s.

For the first 4.1s hare runs:

x=xo+vot+(1/2)*a*t^2

or,x=0+0+(1/2)*1.1*(4.1)^2

or,x=9.2455m

For the next 13.7s hare runs:

x=xo+vt

or,x=9.2455+(4.51*13.7)

x= 71.0325m

A) Equation of position in this part of this race,

x=xo+vot+(1/2)*a*t^2

Where,xo=71.0325

vo= 4.51 m/s^2

The velocity of Hare is zero at position 81 m

Then,v=v0 + at

0=v0+at

or, a=-v0/t

then replacing a=-v0/t in the equation of position and solving for t:

x=x0+v0t+(1/2)(-v0/t)*t^2

x-x0/(1/2v0) =t

Replacing data :

81-71.025/(1/2*4.51) =t

t= 4.42 s

The accelaration is then:

a= - 4.51/4.42 =- 1.02036 m/s^2[Answer]

B) Hare moves 4.1s for accelarating, 13.7 at constant speed, and 4.42s for slowing down.

The Total Time : 4.1+13.7+4.42 = 22.22s [Answer]

C) The tortoise runs 81 m in 22.22s .

The equation for the position can be written as :

x=x0 +v0t+(1/2)at^2

x=0, and v=0 since the tortoise starts from rest .Then solving for a :

2x/t^2 =a

Replacing with data:

2*81/(22.22)^2 = 0.328 m/s^2

a = 0.328 m/s^2.[Answer]