EXERCISE 2.6 A particle follows a planar path defined by x = kč, y = 2k[1 − exp()], such that its speed is v=B², where k and ß are constants. Determine the velocity and acceleration at = 0.5.
EXERCISE 2.6 A particle follows a planar path defined by x = kč, y = 2k[1 − exp()], such that its speed is v=B², where k and ß are constants. Determine the velocity and acceleration at = 0.5.
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![EXERCISE 2.6 A particle follows a planar path defined by x = kč, y = 2k[1 − exp(š)], such that its
speed is v = Bč, where k and ß are constants. Determine the velocity and acceleration at = 0.5.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F780f9839-f737-4aba-91a2-6210989911b1%2F94d5b273-cf7f-4dd9-b9fc-e4ffd4046842%2Fqn6ntbi_processed.png&w=3840&q=75)
Transcribed Image Text:EXERCISE 2.6 A particle follows a planar path defined by x = kč, y = 2k[1 − exp(š)], such that its
speed is v = Bč, where k and ß are constants. Determine the velocity and acceleration at = 0.5.
Expert Solution
Step 1
We have given
x=k and y= 2k [ 1 - e ]
position vector is given by
= xi +yj
=( k )i + 2k [ 1 - e ] j
now velocity ={( k )i + 2k [ 1 - e ] j}
()=ki - 2ke j
()=k( i -2e0.5 j)
V()=
V(0.5)=
V(0.5)=k ..............(1)
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