A car starts from rest and moves with uniform acceleration a on a straight road from time `t=0` to `t=T`. After that, a constant deceleration brings it to rest. In this process the average speed of the car is

Options

A. `(aT)/(4)`
B. `(3aT)/(2)`
C. `(aT)/(2)`
D. `aT`

Explanation

To solve the problem of finding the average speed of the car that starts from rest, accelerates uniformly, and then decelerates to rest, we can follow these steps:

### Step 1: Understand the motion phases
The car undergoes two phases:
1. **Acceleration phase**: From time \( t = 0 \) to \( t = T \) with uniform acceleration \( a \).
2. **Deceleration phase**: From the end of the acceleration phase until it comes to rest.

### Step 2: Calculate the distance during the acceleration phase
Using the equation of motion for distance:
\[ s_1 = ut + \frac{1}{2} a t^2 \]
Since the car starts from rest, \( u = 0 \):
\[ s_1 = 0 + \frac{1}{2} a T^2 = \frac{1}{2} a T^2 \]

### Step 3: Calculate the final velocity at the end of the acceleration phase
Using the equation for final velocity:
\[ v = u + at \]
Again, since \( u = 0 \):
\[ v = 0 + aT = aT \]

### Step 4: Calculate the distance during the deceleration phase
Let the deceleration be \( a_2 \) and the time taken to come to rest be \( t_2 \). The car comes to rest, so the final velocity \( v = 0 \):
Using the equation of motion:
\[ 0 = v - a_2 t_2 \]
Substituting \( v = aT \):
\[ 0 = aT - a_2 t_2 \implies a_2 t_2 = aT \implies t_2 = \frac{aT}{a_2} \]

Now, the distance during the deceleration phase is given by:
\[ s_2 = vt_2 - \frac{1}{2} a_2 t_2^2 \]
Substituting \( v = aT \):
\[ s_2 = aT \cdot t_2 - \frac{1}{2} a_2 t_2^2 \]
Substituting \( t_2 = \frac{aT}{a_2} \):
\[ s_2 = aT \cdot \frac{aT}{a_2} - \frac{1}{2} a_2 \left( \frac{aT}{a_2} \right)^2 \]
\[ s_2 = \frac{a^2 T^2}{a_2} - \frac{1}{2} a_2 \cdot \frac{a^2 T^2}{a_2^2} \]
\[ s_2 = \frac{a^2 T^2}{a_2} - \frac{1}{2} \cdot \frac{a^2 T^2}{a_2} = \frac{1}{2} \cdot \frac{a^2 T^2}{a_2} \]

### Step 5: Calculate total distance and total time
Total distance \( s \):
\[ s = s_1 + s_2 = \frac{1}{2} a T^2 + \frac{1}{2} \cdot \frac{a^2 T^2}{a_2} \]
Total time \( t \):
\[ t = T + t_2 = T + \frac{aT}{a_2} \]

### Step 6: Calculate average speed
Average speed \( v_{avg} \) is given by:
\[ v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{s_1 + s_2}{T + t_2} \]
Substituting the values:
\[ v_{avg} = \frac{\frac{1}{2} a T^2 + \frac{1}{2} \cdot \frac{a^2 T^2}{a_2}}{T + \frac{aT}{a_2}} \]
Factoring out \( \frac{1}{2} T^2 \):
\[ v_{avg} = \frac{\frac{1}{2} T^2 \left( a + \frac{a^2}{a_2} \right)}{T \left( 1 + \frac{a}{a_2} \right)} \]
Simplifying:
\[ v_{avg} = \frac{1}{2} a T \]

### Final Answer
The average speed of the car is:
\[ \boxed{\frac{1}{2} a T} \]