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Shalnov [3]
3 years ago
8

An electric drag racer is much like its piston engine counterpart, but instead it is powered by an electric motor running off of

onboard batteries. These vehicles are capable of covering a 1/4 mile straight-line track in 12 s .
A) Determine the acceleration of the drag racer in units of m/s2. (Assume that the acceleration is constant throughout the race.)
B) How does the value you get compare with the acceleration of gravity?
C) Calculate the final speed of the drag racer in mi/h.
Physics
1 answer:
Lapatulllka [165]3 years ago
6 0

Answer:

5.5879 m/s²

0.56961g

149.9976 mph

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 0.25 miles

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 0.25\times 1609.34=0\times t+\frac{1}{2}\times a\times 12^2\\\Rightarrow a=\dfrac{0.25\times 1609.34\times 2}{12^2}\\\Rightarrow a=5.5879\ m/s^2

The acceleration is 5.5879 m/s²

Dividing by g

\dfrac{a}{g}=\dfrac{5.5879}{9.81}\\\Rightarrow a=0.56961g

The acceleration of these cars is 0.56961 times g

v=u+at\\\Rightarrow v=0+5.5879\times 12\\\Rightarrow v=67.0548\ m/s

Converting mph

67.0548\times \dfrac{3600}{1609.34}=149.9976\ mph

The speed is 149.9976 mph

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Explanation:

Given that,

Mass = 0.254 kg

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Using formula of A and d

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Put the value of \omega=0.628\ rad/s in equation (I) and (II)

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tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

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Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

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From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

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From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

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