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3 years ago

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4) A drag racer starts her car from rest and accelerates at 10.0 m/s² for a distance of 400 m (1/4 mile). (a) How long did it ta

ke the race car to travel this distance? (b) What is the speed of the race car at the end of the run?

1 answer:

mel-nik [20]3 years ago

3 0

Answer:

A) s=1/2at^2

t=√(2s/a)=√(2x400)/10.0)=9.0s

B) v=at

v=10.0x9=90m/s

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Answer:

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Explanation:

Given,

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Direction of the velocity is towards right due to positive velocity.

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Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

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For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

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Given,

Time of impact = t = $25\times 10^{-3}\ sec$

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$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

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$dB = 72.04\: dB$ .

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