To determine how long the car moves uniformly, we can break down the motion into three distinct phases: acceleration, uniform motion, and deceleration. We know the total time of the motion is 25 seconds, and the average velocity is 72 km/h, which we can convert to meters per second for easier calculations.
Step 1: Convert Average Velocity
The average velocity (V_avg) is given as 72 km/h. To convert this to meters per second, we use the conversion factor where 1 km/h is equal to 1/3.6 m/s:
- V_avg = 72 km/h × (1000 m/km) / (3600 s/h) = 20 m/s
Step 2: Calculate Total Distance
Using the average velocity formula, we can find the total distance (D) traveled during the entire motion:
- D = V_avg × total time = 20 m/s × 25 s = 500 m
Step 3: Analyze Each Phase of Motion
Now, let’s analyze each phase of the car's motion:
- Acceleration Phase: The car accelerates from rest (initial velocity, u = 0) with an acceleration (a) of 5 m/s².
- Uniform Motion Phase: The car moves at a constant velocity after the acceleration phase.
- Deceleration Phase: The car decelerates at the same rate (5 m/s²) until it comes to a stop.
Step 4: Calculate Distance During Acceleration and Deceleration
For the acceleration phase, we can use the formula for distance:
- D_acceleration = u × t + 0.5 × a × t²
- Since u = 0, this simplifies to D_acceleration = 0.5 × a × t².
Let t₁ be the time spent accelerating. The distance during acceleration is:
- D_acceleration = 0.5 × 5 m/s² × t₁² = 2.5t₁²
For the deceleration phase, the distance is the same as the acceleration phase because the car decelerates at the same rate:
- D_deceleration = 2.5t₃², where t₃ is the time spent decelerating.
Step 5: Relate Time Variables
The total time of motion is the sum of the times for each phase:
Since the acceleration and deceleration times are equal (t₁ = t₃), we can denote them as t:
- 2t + t₂ = 25 s
- t₂ = 25 s - 2t
Step 6: Calculate Total Distance in Terms of Time
The total distance can also be expressed as the sum of distances from each phase:
- D = D_acceleration + D_uniform + D_deceleration
- D = 2.5t² + V_uniform × t₂ + 2.5t²
- D = 5t² + V_uniform × (25 - 2t)
We need to find V_uniform, which is the final velocity after acceleration:
- V_uniform = u + at = 0 + 5t = 5t
Substituting this into the distance equation gives:
- 500 = 5t² + 5t(25 - 2t)
- 500 = 5t² + 125t - 10t²
- 0 = -5t² + 125t - 500
- 0 = t² - 25t + 100
Step 7: Solve the Quadratic Equation
Using the quadratic formula, t = [ -b ± √(b² - 4ac) ] / 2a:
- a = 1, b = -25, c = 100
- t = [25 ± √(625 - 400)] / 2
- t = [25 ± √225] / 2
- t = [25 ± 15] / 2
This gives us two possible solutions for t:
- t = 20/2 = 10 s
- t = 40/2 = 20 s (not valid since it exceeds total time)
Step 8: Calculate Uniform Motion Time
Now that we have t = 10 s, we can find the time spent in uniform motion:
- t₂ = 25 s - 2t = 25 s - 20 s = 5 s
Therefore, the car moves uniformly for 5 seconds.
