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3 years ago

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a car is traveling at 20 meters/seconds anf is brought to rest by applying breaks over a period of 4 seconds. what is its averag

e acceleration over this time in time interval

1 answer:

allochka39001 [22]3 years ago

8 0

Answer:

<em>2.5m/s²</em>

Explanation:

Given the following:

Initial velocity u = 0m/s

Final velocity v = 20m/s

Time t = 4s

Required

average acceleration over this time in time interval

Using the equation of motion

v = u + at

10 = 0 + a(4)

10 = 4a

a = 10/4

a = 2.5m/s²

<em>Hence its average acceleration over this time in time interval is 2.5m/s²</em>

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Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

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we know that

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The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

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so displacement of the object is given as

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$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

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at t = 2

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Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

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$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

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