Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
(1)
Cheetah speed: 
Its position at time t is given by
(1)
Gazelle speed: 
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
(2)
The cheetah reaches the gazelle when 
. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:
(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah: 
Gazelle: 
So, the gazelle should be ahead of the cheetah of at least
Answer:
Originally : Level = log I / I0
Currently: Level = 10 log I / I0
Level = 10 log 600 = 10 * 2.78 = 27.8
Note the term 1 bel = 10 decibels
