Question

An object starting from rest moves on a straight road for time $t$ and comes to rest finally. The distance is divided into two parts. In the first part, it is accelerated at a constant acceleration and then decelerated at rate $\beta$. The maximum velocity is: C $\quad \frac{\alpha + \beta}{2} \cdot t$

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Answer

To find the maximum velocity of an object that undergoes acceleration and then deceleration, we can follow these steps:

  1. Acceleration Phase:

    • The object starts from rest and accelerates at a constant rate of $\alpha$ for time $ t_1 $.
    • The velocity $ V $ at the end of this phase can be expressed using the equation of motion: $$ V = 0 + \alpha t_1 \quad \text{(i)} $$
  2. Deceleration Phase:

    • After reaching the maximum velocity $ V $, the object decelerates at a constant rate $\beta$ for time $ t_2 $.
    • The final velocity at the end of this phase is zero: $$ 0 = V - \beta t_2 $$
    • From equation (i), since $ V = \alpha t_1 $, we can substitute this in: $$ 0 = \alpha t_1 - \beta t_2 \quad \text{(ii)} $$
  3. Total Time:

    • The total time $ t $ is the sum of $ t_1 $ and $ t_2 $: $$ t_1 + t_2 = t $$
  4. Solving for Maximum Velocity:

    • From equation (ii), solve for $ t_2 $: $$ t_2 = \frac{\alpha t_1}{\beta} $$

    • Substitute $ t_2 $ into the total time equation: $$ t_1 + \frac{\alpha t_1}{\beta} = t $$

    • After rearranging: $$ t_1 \left(1 + \frac{\alpha}{\beta} \right) = t $$

    • Solve for $ t_1 $: $$ t_1 = \frac{t}{1 + \frac{\alpha}{\beta}} = \frac{t \beta}{\alpha + \beta} $$

    • Then, substituting $ t_1 $ back into equation (i) to find $ V $: $$ V = \alpha t_1 = \alpha \left(\frac{t \beta}{\alpha + \beta}\right) = \frac{\alpha \beta t}{\alpha + \beta} $$

Thus, the maximum velocity $ V $ is: $$ \boxed{\frac{\alpha \beta t}{\alpha + \beta}} $$

Therefore, the correctly framed answer is indeed from option D: $$ \frac{\alpha \beta}{\alpha + \beta} \cdot t $$


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