Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
X =(3.00x4.00 x3-1.00t x 2.00) x m
x= (12.00x3- 1.00 x2.00) x m
x= 36.00 -1.00 x 2.00) x m
x = (36.00 -2.00) x m
x =( 34.00) x m
x =34.00 times m
The simplest way to do this is to set up equivalent fractions, like this-

=

Solve for x by using cross multiplication.
40*2.2= 88
1*x=88
x=88
Therefore, the boy weighs 88lbs.