Question:medium
The acceleration \( f \) (in ft/sec\(^2\)) of a particle after a time \( t \) seconds starting from rest is given by:
\[ f = 6 - \sqrt{1.2t}. \]
Then the maximum velocity \( v \) and the time \( T \) to attain this velocity are:
The acceleration \( f \) (in ft/sec\(^2\)) of a particle after a time \( t \) seconds starting from rest is given by:
\[ f = 6 - \sqrt{1.2t}. \]
Then the maximum velocity \( v \) and the time \( T \) to attain this velocity are:
\[ f = 6 - \sqrt{1.2t}. \]
Then the maximum velocity \( v \) and the time \( T \) to attain this velocity are:
Show Hint
To find maximum velocity, differentiate the velocity expression and set the derivative equal to zero. Integrate acceleration for the velocity function
Updated On: Nov 28, 2025
- \( T = 20 \, \text{sec} \)
- \( v = 60 \, \text{ft/sec} \)
- \( T = 30 \, \text{sec} \)
- \( v = 40 \, \text{ft/sec} \)
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