A tortoise and hare start from rest and have a race. As the race begins, both ac

Question

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 11.7 seconds, before getting tired and slowing down with constant acceleration coming to rest 72 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.

1)

How fast is the hare going 1.9 seconds after it starts?

2)

How fast is the hare going 12.2 seconds after it starts?

3)

How far does the hare travel before it begins to slow down?

4)

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

5)

What is the total time the hare is moving?

6)

What is the acceleration of the tortoise?

Explanation / Answer

for the hare :

upto 4.7s u=0 a=0.9m/s => v=u+at = 0.9(4.7)=4.23 m/s.

=> s=ut+1/2(a(t^2)=0+(0.5 x 0.9 x 4.7^2)

=9.9405 m.

for 11.7 s v is constant. => v= 4.23 m/s. s=vt=4.23(11.7)=49.491 m

after 11.7+4.7=16.4 s it retards and comes to rest.

distance left=72-9.9405-49.491=12.5685 m

u=4.23m/s. v=0. v^2-u^2=2as => 0 -4.23^2 = 2 x a x 12.5685

=> a=-0.712 m/s^2

v=u+at => 0=4.23-(0.712)(t) => t= 5.94 s.

total time = 16.4 +5.94 = 22.34 s.

tortoise acc. uniformly from rest. s=72m t=22.34 s. u=0

s=ut+0.5at^2 => 72 = 0+(0.5)(a)(22.34^2)

=> a = 0.2885 m/s^2

a)upto 4.7s v=0.9t =0.9 x 1.9 = 1.71 m/s.

b)after 4.7s it travels with same speed =4.23 m/s

c)before slowing down ,distance =9.9405+49.491=59.8505 m

d) retardation of hare=0.712 m/s^2

e)total time =22.34s

f)acc. of tortoise = 0.2885 m/s^2

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