A scooter accelerates from rest for time `t_(1)` at constant rate `a_(1)` and then retards at constant rate `a_(2)` for time `t_(2)` and comes to rest. The correct value of `(t_(1))/(t_(2))` will be :-

Options

A. `(a_(1)+a_(2))/(a_(2))`
B. `(a_(2))/(a_(1))`
C. `(a_(1))/(a_(2))`
D. `(a_(1)+a_(2))/(a_(1))`

Answer

To solve the problem, we need to analyze the motion of the scooter during both the acceleration and deceleration phases. Let's break it down step by step.

### Step 1: Determine the final velocity after acceleration
The scooter starts from rest and accelerates at a constant rate \( a_1 \) for a time \( t_1 \). Using the first equation of motion:

\[ v_1 = u + a_1 t_1 \]

Since the initial velocity \( u = 0 \):

\[ v_1 = a_1 t_1 \quad \text{(Equation 1)} \]

### Step 2: Determine the relationship during deceleration
After reaching velocity \( v_1 \), the scooter then decelerates at a constant rate \( a_2 \) for a time \( t_2 \) until it comes to rest. Using the second equation of motion:

\[ 0 = v_1 - a_2 t_2 \]

Rearranging gives:

\[ v_1 = a_2 t_2 \quad \text{(Equation 2)} \]

### Step 3: Set the equations equal to each other
From Equations 1 and 2, we have:

\[ a_1 t_1 = a_2 t_2 \]

### Step 4: Solve for the ratio \( \frac{t_1}{t_2} \)
To find the ratio \( \frac{t_1}{t_2} \), we can rearrange the equation:

\[ \frac{t_1}{t_2} = \frac{a_2}{a_1} \]

This gives us the required ratio of the times.

### Conclusion
The correct value of \( \frac{t_1}{t_2} \) is:

\[ \frac{t_1}{t_2} = \frac{a_2}{a_1} \]