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Nataly [62]
3 years ago
8

A particle moves along the x axis. It is intially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.32

0 m/s^2. Suppose it moves as a particle under constant acceleration for 4.50m/s^2. Find its position.
Physics
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

The position of the particle is -2.34 m.

Explanation:

Hi there!

The equation of position of a particle moving in a straight line with constant acceleration is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the particle at a time t:

x0 = initial position.

v0 = initial velocity.

t = time

a = acceleration

We have the following information:

x0 = 0.270 m

v0 = 0.140 m/s

a = -0.320 m/s²

t = 4.50 s  (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).

Then, we have all the needed data to calculate the position of the particle:

x = x0 + v0 · t + 1/2 · a · t²

x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²

x = -2.34 m

The position of the particle is -2.34 m.

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<h3>What is the Kichoff's loop rule?</h3>

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In a series RLCcircuit, the voltages are not added by scalar addition but by vector addition.

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