A car starts from rest and moves with a constant acceleration of 5 𝑚/𝑠2 for 10 seconds before the driver applies the break.  It then decelerates for 5 seconds coming to rest.  The average speed of the car over the entire journey is

To solve this problem, we need to break it down into two parts: the acceleration phase and the deceleration phase.

1. Acceleration Phase: The car starts from rest and moves with a constant acceleration of 5 m/s² for 10 seconds. We can use the formula for final velocity in uniformly accelerated motion, which is v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Since the car starts from rest, u = 0. So, v = 0 + 5*10 = 50 m/s. This is the final velocity of the car at the end of the acceleration phase.

The distance covered in this phase can be calculated using the formula s = ut + 0.5at². Again, since u = 0, s = 0.5510² = 250 m.

2. Deceleration Phase: The car then decelerates for 5 seconds coming to rest. This means the final velocity is 0. We can use the formula v = u - at, where now u is the initial velocity at the start of this phase (which is the final velocity of the acceleration phase), a is the deceleration, and t is the time. We know that the car comes to rest, so v = 0. Therefore, 0 = 50 - a*5. Solving for a, we get a = 50/5 = 10 m/s².

The distance covered in this phase can be calculated using the formula s = ut - 0.5at². So, s = 505 - 0.510*5² = 125 m.

3. Average Speed: The total distance covered is the sum of the distances covered in the acceleration and deceleration phases, which is 250 m + 125 m = 375 m. The total time is the sum of the times of the two phases, which is 10 s + 5 s = 15 s.

The average speed is the total distance divided by the total time, which is 375 m / 15 s = 25 m/s. So, the average speed of the car over the entire journey is 25 m/s.

To solve this problem, we need to break it down into two parts: the acceleration phase and the deceleration phase.

1. Acceleration Phase: The car starts from rest and moves with a constant acceleration of 5 m/s² for 10 seconds. We can use the formula for final velocity in uniformly accelerated motion, which is v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Since the car starts from rest, u = 0. So, v = 0 + 5*10 = 50 m/s. This is the final velocity of the car at the end of the acceleration phase.

The distance covered in this phase can be calculated using the formula s = ut + 0.5at². Again, since the car starts from rest, u = 0. So, s = 0 + 0.5510² = 250 m.

2. Deceleration Phase: The car decelerates for 5 seconds coming to rest. This means the initial velocity at the start of this phase is the final velocity from the acceleration phase, which is 50 m/s. The final velocity is 0 m/s.

The distance covered in this phase can be calculated using the formula s = ut - 0.5at². Here, a is the deceleration, which is equal to the negative of the acceleration. So, s = 505 - 0.55*5² = 125 m.

3. Average Speed: The total distance covered is the sum of the distances covered in the acceleration and deceleration phases, which is 250 m + 125 m = 375 m. The total time taken is the sum of the times taken in the acceleration and deceleration phases, which is 10 s + 5 s = 15 s.

The average speed is the total distance covered divided by the total time taken, which is 375 m / 15 s = 25 m/s.

So, the average speed of the car over the entire journey is 25 m/s.