To solve this, let's break the motion into three phases:

1. Acceleration phase: The car accelerates from rest.
2. Constant speed phase: The car travels at a constant speed.
3. Deceleration phase: The car decelerates uniformly until it stops.

Phase 1: Acceleration

• Initial velocity ($u_1$) = 0 m/s (starts from rest)
• Final velocity ($v_1$) = 26.0 m/s
• Acceleration ($a$) = 2.00 m/s²
• Time taken ($t_1$) = ?

We use the formula: $v_1 = u_1 + a t_1$ $t_1 = \frac{v_1 - u_1}{a} = \frac{26.0 \, \text{m/s} - 0 \, \text{m/s}}{2.00 \, \text{m/s}^2} = 13.0 \, \text{s}$

Phase 2: Constant Speed

• Speed during this phase ($v_2$) = 26.0 m/s
• Time taken ($t_2$) = 51.0 s

Phase 3: Deceleration

• Initial velocity ($u_3$) = 26.0 m/s
• Final velocity ($v_3$) = 0 m/s (comes to a stop)
• Time taken ($t_3$) = 5.00 s
• Deceleration ($a_3$) = ?

We use the formula: $v_3 = u_3 + a_3 t_3$ Solving for $a_3$: $0 = 26.0 \, \text{m/s} + a_3 \cdot 5.00 \, \text{s}$ $a_3 = - \frac{26.0}{5.00} = -5.20 \, \text{m/s}^2$ (Negative sign indicates deceleration)

Total time in motion

The total time the car is in motion is the sum of the time in all three phases: $t_{\text{total}} = t_1 + t_2 + t_3 = 13.0 \, \text{s} + 51.0 \, \text{s} + 5.00 \, \text{s} = 69.0 \, \text{s}$

Average velocity

To find the average velocity, we need to find the total distance traveled and divide it by the total time.

Distance in phase 1 (acceleration):

Using the equation: $d_1 = u_1 t_1 + \frac{1}{2} a t_1^2$ $d_1 = 0 + \frac{1}{2} \cdot 2.00 \, \text{m/s}^2 \cdot (13.0 \, \text{s})^2 = 169 \, \text{m}$

Distance in phase 2 (constant speed):

$d_2 = v_2 \cdot t_2 = 26.0 \, \text{m/s} \cdot 51.0 \, \text{s} = 1326 \, \text{m}$

Distance in phase 3 (deceleration):

Using the equation: $d_3 = u_3 t_3 + \frac{1}{2} a_3 t_3^2$ $d_3 = 26.0 \, \text{m/s} \cdot 5.00 \, \text{s} + \frac{1}{2} \cdot (-5.20 \, \text{m/s}^2) \cdot (5.00 \, \text{s})^2 = 65 \, \text{m} + (-65 \, \text{m}) = 65 \, \text{m}$

Total distance:

$d_{\text{total}} = d_1 + d_2 + d_3 = 169 \, \text{m} + 1326 \, \text{m} + 65 \, \text{m} = 1560 \, \text{m}$

Average velocity:

$v_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{1560 \, \text{m}}{69.0 \, \text{s}} \approx 22.6 \, \text{m/s}$

Final answers:

• Total time in motion: $69.0 \, \text{s}$
• Average velocity: $22.6 \, \text{m/s}$

Would you like more details or clarifications?

Here are some related questions for further exploration:

1. How far does the car travel during the acceleration phase alone?
2. What would the stopping distance be if the deceleration was doubled?
3. How would the average velocity change if the constant speed phase lasted longer?
4. What is the total deceleration experienced during the stopping phase?
5. How would the total time change if the car accelerated at a faster rate?

Tip: When solving motion problems, always break down the problem into different phases and apply the right kinematic equations for each phase.