12) A particle moves in a straight line, the algebraic measure of its velocity \( v=5+4 t-t^{2} \) then it reaches to maximum speed at \( t=\ldots \ldots \ldots \) during \( t \in[0,6] \) \( \begin{array}{llll}\text { (a) } 3 & \text { (b) } 1 & \text { (c) } 2 & \text { (d) } 5 \text { or } 1\end{array} \)

To find the time \( t \) at which the particle reaches its maximum speed, we need to analyze the velocity function given by: \[ v(t) = 5 + 4t - t^2 \] ### Step 1: Find the critical points To find the maximum speed, we first need to find the critical points of the velocity function. This is done by taking the derivative of \( v(t) \) and setting it to zero: \[ v'(t) = \frac{d}{dt}(5 + 4t - t^2) \] ### Step 2: Calculate the derivative Let's calculate the derivative: \[ v'(t) = 4 - 2t \] ### Step 3: Set the derivative to zero Now, we set the derivative equal to zero to find the critical points: \[ 4 - 2t = 0 \] ### Step 4: Solve for \( t \) Solving for \( t \): \[ 2t = 4 \implies t = 2 \] ### Step 5: Check the interval We need to check if this critical point \( t = 2 \) is within the interval \( [0, 6] \). Since \( 2 \) is within this interval, we will evaluate the velocity at this point. ### Step 6: Evaluate the velocity at the endpoints and the critical point We will evaluate \( v(t) \) at \( t = 0 \), \( t = 2 \), and \( t = 6 \): 1. \( v(0) = 5 + 4(0) - (0)^2 = 5 \) 2. \( v(2) = 5 + 4(2) - (2)^2 = 5 + 8 - 4 = 9 \) 3. \( v(6) = 5 + 4(6) - (6)^2 = 5 + 24 - 36 = -7 \) ### Step 7: Determine the maximum speed From the evaluations: - \( v(0) = 5 \) - \( v(2) = 9 \) - \( v(6) = -7 \) The maximum speed occurs at \( t = 2 \). ### Conclusion The particle reaches its maximum speed at \( t = 2 \). Thus, the answer is \( \text{(c) } 2 \).