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  (alpha + beta / 2) * t
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 $\quad \frac{\alpha + \beta}{2} \cdot t$
✨ Free AI Tools for You
Powered by Chatterbot AI
Answer
To find the maximum velocity of an object that undergoes acceleration and then deceleration, we can follow these steps:

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: $$ V = 0 + \alpha t_1 \quad \text{(i)} $$

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: $$ 0 = \alpha t_1  \beta t_2 \quad \text{(ii)} $$

Total Time:
 The total time $ t $ is the sum of $ t_1 $ and $ t_2 $: $$ t_1 + t_2 = t $$

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: $$ \frac{\alpha \beta}{\alpha + \beta} \cdot t $$
India's 1^{st} AI Doubt Solver for CBSE, JEE, and NEET
Ask a Question for Freeand then it's just ₹212 a month