Starting from rest, a car accelerates down a straight road with constant acceleration a for a time t,

then the direction of the acceleration is reversed, (i.e., it is a), and the car comes to a stop in an
additional time t, the time for the whole trip being 2t. At what time, or times, is the average
velocity of the car for the trip equal to its instantaneous velocity during the trip?

To find the time or times at which the average velocity of the car is equal to its instantaneous velocity during the trip, we can set up the equations for average velocity and instantaneous velocity and solve for the time(s) that satisfy the condition.

First, let's define the terms:
- Initial velocity = 0 (since the car starts from rest)
- Acceleration = a (constant)
- Time taken for the first part of the trip (acceleration phase) = t
- Time taken for the second part of the trip (deceleration phase) = t
- Total time for the whole trip = 2t

Average Velocity:
Average velocity is given by the formula:
Average velocity = Total displacement / Total time

Instantaneous Velocity:
Instantaneous velocity is the velocity at a particular instant in time. It can be found using the formula for velocity:
Instantaneous velocity = Initial velocity + (Acceleration × Time)

Now, let's calculate the average and instantaneous velocities separately for the two parts of the trip.

PART 1: Acceleration Phase
During the acceleration phase, the car starts from rest and accelerates with constant acceleration a for a time t.
Average velocity of the first part:
Average velocity = Total displacement / Total time
As the car starts from rest, the average velocity for the first part will be half of its final velocity at time t.
Average velocity = (1/2) × Final velocity (at time t)

Instantaneous velocity of the first part:
Instantaneous velocity = Initial velocity + (Acceleration × Time)
Instantaneous velocity = 0 + (a × t) = a × t

PART 2: Deceleration Phase
During the deceleration phase, the car comes to a stop in an additional time t by reversing the direction of acceleration (i.e., -a).
Average velocity of the second part:
Average velocity = Total displacement / Total time
As the car starts from rest, the average velocity for the second part will be half of its final velocity at time 2t.
Average velocity = (1/2) × Final velocity (at time 2t)

Instantaneous velocity of the second part:
Instantaneous velocity = Initial velocity + (Acceleration × Time)
Since the acceleration direction is -a, the initial velocity for the second part will be -a × t.
Instantaneous velocity = -a × t + (-a × t) = -2a × t

Now, set the average velocity equal to the instantaneous velocity for both parts of the trip and solve for time:
For the acceleration phase:
(1/2) × Final velocity (at time t) = a × t

For the deceleration phase:
(1/2) × Final velocity (at time 2t) = -2a × t

Solve these two equations to find the values of t that satisfy the conditions. Substituting the equations for final velocity:
(1/2) × (0 + a × t) = a × t
(1/2) × (-a × t + 0) = -2a × t

Simplifying these equations, we get:
(1/2) × a × t = a × t
(-1/2) × a × t = -2a × t

Canceling out the common factors and rearranging the equations, we obtain:
1/2 = 1
-1/2 = -2

However, these equations are not possible. The average velocity can never be equal to the instantaneous velocity for this situation.

Therefore, there is no time or times at which the average velocity of the car for the entire trip is equal to its instantaneous velocity.