Question #128853

A particle travels along a straight line with a velocity of v= (20 − 0.05s^2) m/s, where s is in meters. Determine the acceleration of the particle at s = 15 m.

Expert's answer

Velocity v=(20-0.05s"^2" )m/s

at s=15m ,

velocity v=(20-0.05"\\times 15^2)"

v=18.75m/s

as we know acceleration is the rate of change of velocity

a="\\frac{dv}{dt}"

="\\frac{d(20-0.05s^2)}{dt}"

= 0-0.05"\\times 2s\\times \\frac{ds}{dt}"

=-0.05"\\times 2 \\times 15\\times v" ( since velocity is rate of change of distance)

= "-1.5\\times18.75"

= - 28.125m/s"^2"

Hence the acceleration at s=15m is - 28.125m/s"^2"

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