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1 year ago

If the car is traveling at a velocity of 15 m/s, what is the approximate centripetal acceleration of the car?.

Physics

1 answer:

BaLLatris [955]1 year ago

7 0

The vehicle's centripetal acceleration is equal to 22.5m/s²

Radius, r = 10 meter

Speed, V = 15 m/s

To ascertain the car's centripetal acceleration

A(c) = V²/R

We obtain the following when we enter the formula's parameters:

A(c) = 152/10

A(c) = 225/10

A(c) = 22.5m/s²

<h3>What is Centripetal acceleration ?</h3>

When an item moves in a circular route, one of its motion characteristics is centripetal acceleration. Any motion in a circle with an acceleration vector pointing in the direction of the circle's centre is referred to as centripetal acceleration.

Centripetal forces cause accelerations at the centripetal axis. With the exception of the Earth's rotation around the Sun, any satellite's circular motion around a celestial body is brought on by the centripetal force produced by their mutual gravitational pull.

Hence, Centripetal acceleration is

22.5 m/s²

Learn more about Centripetal acceleration here:

brainly.com/question/79801

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A train travelling at 20m/s undergoes a uniform deceleration of 2m/s2 when brakes are applied.. calculate the time taken to come

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Answer:

a) 10 seconds

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Explanation:

<u>Accelerated Motion </u>

If an object is moving at contant speed, we know it has zero net force applied. As soon as an unbalanced force appears, the object accelerates (positive or negative) depending on the direction of the force with respect to movement. The speed of an object subject to acceleration is given by

$v_f=v_o+at$

The distance traveled is given by

$\displaystyle x=v_ot+\frac{at^2}{2}$

Our train is travelling at 20m/s when it decelerates at $-2\ m/s^2$ (negative for it's against movement). If nothing else happens, it will eventually stop ( $v_f=0$ ). We can compute the time at which it happens :

$\displaystyle t=\frac{v_f-v_o}{a}$

$Knowing\ v_f=0$

$\displaystyle t=\frac{-20}{-2}=10\ sec$

The distance traveled is

$\displaystyle x=v_ot+\frac{at^2}{2}$

$\displaystyle x=(20)(10)-\frac{2(10)^2}{2}$

$\displaystyle x=100\ m$

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