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2 years ago

While traveling along a highway, a driver slows from 27 m/s to 13 m/s in 11 seconds. What is the automobile’s acceleration?

Physics

1 answer:

Travka [436]2 years ago

4 0

Answer:

$a=-1.27\ m/s^2$

Explanation:

Given that,

The initial speed of a driver, u = 27 m/s

Final speed of the driver, v = 13 m/s

Time, t = 11 s

We need to find the automobile's acceleration. We know that the acceleration of an object is equal to the change in speed divided by time taken.

$a=\dfrac{13-27}{11}\\\\a=-1.27\ m/s^2$

So, the magnitude of automobile's acceleration is $1.27\ m/s^2$ .

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What makes speed, velocity, acceleration, distance, and time different?

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2 years ago

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2 years ago

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Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, $\omega_{rotation}$ = Angle turned/Time

∴ $\omega_{rotation}$ = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

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Part B

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Therefore, the average angular speed of the planet around its neighboring star, $\omega _{Star}$ , is given as follows;

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2 years ago