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 a car that starts from rest, accelerates uniformly, and then decelerates to rest, we can break the solution down into clear steps.

### Step-by-Step Solution:

1. **Understanding the Motion Phases**:
- The car starts from rest (initial velocity \( u = 0 \)).
- It accelerates uniformly with acceleration \( a \) for a time \( T \).
- After time \( T \), it decelerates to rest with a constant deceleration.

2. **Calculating Distance during Acceleration**:
- The distance \( s_1 \) covered during the acceleration phase can be calculated using the formula:
\[ s_1 = ut + \frac{1}{2} a t^2 \]
- Since \( u = 0 \), this simplifies to:
\[ s_1 = \frac{1}{2} a T^2 \]

3. **Finding Final Velocity at the End of Acceleration**:
- The final velocity \( v \) at the end of the acceleration phase can be calculated using:
\[ v = u + aT \]
- Again, since \( u = 0 \):
\[ v = aT \]

4. **Calculating Distance during Deceleration**:
- Let the time taken to come to rest from velocity \( v \) be \( t_2 \).
- The distance \( s_2 \) covered during deceleration can be calculated using:
\[ s_2 = vt_2 - \frac{1}{2} a_2 t_2^2 \]
- Here, \( a_2 \) is the deceleration, and since the car comes to rest, we can express \( a_2 \) as \( a_2 = \frac{v}{t_2} = \frac{aT}{t_2} \).
- Substituting \( v \) into the distance formula gives:
\[ s_2 = (aT)t_2 - \frac{1}{2} \left(\frac{aT}{t_2}\right) t_2^2 \]
- Simplifying this yields:
\[ s_2 = aTt_2 - \frac{1}{2} aTt_2 = \frac{1}{2} aTt_2 \]

5. **Total Distance and Total Time**:
- The total distance \( S \) is:
\[ S = s_1 + s_2 = \frac{1}{2} a T^2 + \frac{1}{2} a T t_2 \]
- The total time \( T_{total} \) is:
\[ T_{total} = T + t_2 \]

6. **Calculating Average Speed**:
- The average speed \( v_{avg} \) is given by:
\[ v_{avg} = \frac{S}{T_{total}} = \frac{\frac{1}{2} a T^2 + \frac{1}{2} a T t_2}{T + t_2} \]
- Factoring out \( \frac{1}{2} a \):
\[ v_{avg} = \frac{\frac{1}{2} a (T^2 + T t_2)}{T + t_2} \]

7. **Final Simplification**:
- The average speed can be simplified further depending on the relationship between \( T \) and \( t_2 \). However, for uniform acceleration and deceleration, the average speed can be approximated to:
\[ v_{avg} = \frac{1}{2} a T \]

### Final Answer:
The average speed of the car during the entire motion is:
\[ v_{avg} = \frac{1}{2} a T \]