Answer
By second equation of motion
"s=ut+\\frac{1}{2}at^2\\\\11=0(5) +\\frac{1}{2}a(5) ^2\\\\a=0.88m\/s^2"
Velocity after first 5second
"v=0.88(5) =4.4m\/s"
When force will removed then acceleration will be zero
So distance travelled in next 5 second
"s=4.4\\times5=22m"
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