if t^2+100=4vt then find the possible values of v

mechanics

• Bhavya
• 3156 दृश्य
• 1 उत्तर

• We solve this equation for v:br /br /4vt=100+t^2, v=(100+t^2)/4t.br /br /Now, to the possible values.br /br /We cant calculate v when t=0 (divizion by 0). Also, t can not be negative, so v too. Then we find the minimum for v(t).br /br /br /br /v^2;(t)=-25/t^2+1/4=0 so t=10 is a local minimum.br /br /Minimal value for v is v(10)=(100+100)/40=5m/s maximum is any.br /br /So the answer - v is equal or greater than 5 m/s.

  • Shruti
  • 02 अप्रैल
  • 0 टिप्पणी

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