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A particle is traveling along the parabolic path y = 0.25x2. If x = (2t2) m, where t is in...

A particle is traveling along the parabolic path y = 0.25x2. If x = (2t2) m, where t is in seconds, determine the magnitude of the particle’s velocity and acceleration when t = 2 s.

Step-by-Step Solution

Solution 1
Step #1 of 8

Component of the particle position is

Given

Velocity of the particle along the axis is expressed as

Accordingly, we have from

So, we have

Step #2 of 8

Velocity of the particle along the axis is expressed as

Given

By differentiating the above the relation, we have

Step #3 of 8

By substituting and the value of into the above relation, we have

Step #4 of 8

When:

Step #5 of 8

Magnitude of the particle velocity at is expressed as

By substituting the values of the parameters into the above relation, we have

Magnitude of the particle velocity at is

Step #6 of 8

Acceleration of the particle along the axis is expressed as

Accordingly, we have from

Step #7 of 8

Acceleration of the particle along the axis is expressed as

Accordingly, we have from

Step #8 of 8

When:

Magnitude of the particle acceleration at is expressed as

By substituting the values of the parameters into the above relation, we have

Magnitude of the particle acceleration at is

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