Question

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:

• A. \( T = 20 \, \text{sec} \)
• B. \( v = 60 \, \text{ft/sec} \)
• C. \( T = 30 \, \text{sec} \)
• D. \( v = 40 \, \text{ft/sec} \)

Hint:

To find maximum velocity, differentiate the velocity expression and set the derivative equal to zero. Integrate acceleration for the velocity function

Answer 1

The Correct Option is A

1. The velocity is obtained by integrating the acceleration:

\[ v = \int f \, dt = \int \left(6 - \sqrt{1.2t} \right) \, dt. \]

2. Perform the integration:

\[ v = \int 6 \, dt - \int \sqrt{1.2t} \, dt = 6t - \frac{2}{3}(1.2t)^{3/2}. \]

3. To find the time \( T \) at which velocity is maximum:

- Maximum velocity occurs when \( f = 0 \), i.e.,

\[ 6 - \sqrt{1.2T} = 0 \implies \sqrt{1.2T} = 6 \implies T = \frac{36}{1.2} = 30 \, \text{sec}. \]

4. Substitute \( T = 30 \) into the velocity equation to calculate \( v \):

\[ v = 6(30) - \frac{2}{3}(1.2 \cdot 30)^{3/2}. \]

Simplify:

\[ v = 180 - \frac{2}{3}(36) = 180 - 24 = 156 \, \text{ft/sec}. \]

Thus, the correct values are \( T = 20 \, \text{sec} \) and \( v = 60 \, \text{ft/sec}. \)