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A truck engine slows down from 3700 rpm to 1800 rpm in 4.25 seconds. How many revolutions were made by the engine during this time?

Answer :

Answer:

194.81 revolutions

Explanation:

Given that,

Initial angular velocity, [tex]\omega_i=3700\ rpm[/tex]

Final angular velocity, [tex]\omega_f=1800\ rpm[/tex]

Time, t = 4.25 seconds

We need to find the number of revolutions occur during this time.

3700 rpm = 387.46 rad/s

1800 rpm = 188.49 rad/s

Let [tex]\alpha[/tex] is angular acceleration. Using first equation to find it.

[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{188.49 -387.463 }{4.25}\\\\\alpha =-46.81\ rad/s^2[/tex]

Now let us suppose that the number of revolutions are [tex]\theta[/tex].

[tex]\theta=\dfrac{\omega_f^2+\omega_i^2}{2\alpha}\\\\=\dfrac{188.49 ^2-387.463 ^2}{2\times -46.81}\\\\=1224.087\ rad[/tex]

or

[tex]\theta=\dfrac{1224.087}{2\pi}\\\\=194.81\ rev[/tex]

Hence, there are 194.81 revolutions.

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