Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)
(a) 
The frequency of a wave is given by:

where
v is the wave's speed
is the wavelength
For the red laser light in this problem, we have
(speed of light)

Substituting,

(b) 427.6 nm
The wavelength of the wave in the glass is given by

where
is the original wavelength of the wave in air
n = 1.48 is the refractive index of glass
Substituting into the formula,

(c) 
The speed of the wave in the glass is given by

where
is the original speed of the wave in air
n = 1.48 is the refractive index of glass
Substituting into the formula,

Answer:
1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.
2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.
3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.
The gravitional potential energy, relative to the bottom of the giant drop, in joules, is (9800) times (the height of the drop in meters).
That's the PE of the empty car only, not counting any hapless screaming souls who may be trapped in it at that moment.