Respuesta :
Answer:
acceleration is 10.05 m/s² or 32.97 ft/s²
distance is 40.22 m or 131.95 ft
Explanation:
given data
velocity = 45 mph = 20.1168 m/s
time = 2 sec
to find out
acceleration and distance
solution
first we will apply here formula for acceleration that is
v = u +at ........................1
here v is velocity and u is initial speed and t is time and a is acceleration
so put here all value
20.11 = 0 + a(2)
a = 10.05 m/s
so acceleration is 10.05 m/s² or 32.97 ft/s²
and
distance is calculated as
distance = v × t ..................2
put here value
distance = 20.11 × 2
distance = 40.22
so distance is 40.22 m or 131.95 ft
The acceleration of the cheetah in (ft/s²) is 33 ft/s²
The acceleration of the cheetah in (m/s²) is 10.1 m/s²
The distance traveled by the cheetah in feet is 66 ft
The distance traveled by the cheetah in meet is 20 m
The given parameter;
speed of the cheetah, v = 45 mph
The acceleration of the cheetah in (ft/s²) is calculated as;
5280 ft = 1 mile
[tex]a = \frac{v}{t} = v \times \frac{1}{t} = \frac{45 \ miles}{hr} \times \frac{5280 \ ft}{1 \ mile} \times \frac{1 \ hr}{3600 \ s} \times \frac{1}{2\ s} = 33 \ ft/s^2[/tex]
The acceleration of the cheetah in (m/s²) is calculated as;
1609 m = 1 mile
[tex]a = \frac{v}{t} = v \times \frac{1}{t} = \frac{45 \ miles}{hr} \times \frac{1609 \ m}{1 \ mile} \times \frac{1 \ hr}{3600 \ s} \times \frac{1}{2\ s} = 10.1 \ m/s^2[/tex]
The distance traveled by the cheetah in feet;
[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2as\\\\2as = v^2 \\\\s = \frac{v^2}{2a} \\\\s = \frac{(66)^2}{2(33)} \\\\s = 66 \ ft[/tex]
The distance traveled by the cheetah in meet:
[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2as\\\\2as = v^2 \\\\s = \frac{v^2}{2a} \\\\s = \frac{(20.1)^2}{2(10.1)} \\\\s = 20 \ m[/tex]
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