totoise anc hare start from rest and have a race. As the race begins, both accel

Question

totoise anc hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 1.4 m/s2 for 4.8 seconds. It then continues at a constant speed for 13.8 seconds before getting tired and slowing down with constant acceleration coming to rest 117 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a Siup 1) How fast is the hare going 1.9 seconds after it starts? See solution 2) How fast is the hare going 8.8 seconds after it starts? See solution 3) How far does the hare travel before it begins to slow downi? See solution Sabit 4) What is the acceleration of the hare once it begins to slow down? See solution m/s2 5) What is the total time the hare is moving? See solution

Explanation / Answer

1)
for hare:
vf =?
vi = 0 m/s
a= 1.4 m/s^2
t=1.9 s
use:
vf = vi + a*t
= 0 + 1.4*1.9
= 2.66 m/s
Answer: 2.66 m/s

2)
speed is same after 4.8 s
for hare:
vf =?
vi = 0 m/s
a= 1.4 m/s^2
t=4.8 s
use:
vf = vi + a*t
= 0 + 1.4*4.8
= 6.72 m/s
Answer: 6.72 m/s

3)
for 1st 4.8 s,
d1=?
vi = 0 m/s
a= 1.4 m/s^2
t=4.8 s
use:
d1 = vi*t + 0.5*a*t^2
= 0 + 0.5*1.4*4.8^2
= 16.13 m

for 4.8 s to 13.8 s
d2=?
t = 13.8-4.8 = 9 s
v = 6.72 m/s
d2=v*t
=6.72*9
=60.48 m

Total distance,
d = d1+d2
= 16.13 + 60.48
= 76.61 m
Answer: 76.61 m

4)
while slowing down,
d = 117-76.61 = 40.39 m
vf= 0
vi = 6.72 m/s

use:
vf^2 = vi^2 + 2*a*d
0 = 6.72^2 + 2*a*40.39
a = -0.56 m/s^2
Answer: -0.56 m/s^2

I am allowed to answer only 4 parts at a time

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