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A particle is moving along the x-axis. Its position at time t>0 is e^(2-t). What is...
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the position in the x direction is given by,
`x = e^((2-t)) --- (1)`
` ` now the acceleration in the x direction is given by, ` ``(d^2x)/dt^2`
lets differenciate the equation (1) to obtain an expression for the acceleration
`x = e^((2-t))`
`dx/dt = -e^((2-t))`
`(d^2x)/dt^2 = e^((2-t))`
` ` when t = 2,` `
`(d^2x)/dt^2 = e^((2-t)) = e^((2-2)) = e^0 = 1`
` <br> `
Hence the acceleration when t = 2 is 1
Posted by malagala on May 23, 2012 at 4:39 AM (Answer #1)
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