A car starts moving along a line, first with acceleration a=5 `ms^(-2)` starting from rest then uniformly and finally decelerating at the same rate a, comes to rest.The total time of motion is `tau=25 s`. The average velocity during the time is equal to lt v gt =72 km/hr.How long does the partial move uniformly ?

Options

A. 10 s
B. 12 s
C. 20 s
D. 15 s

Explanation

To solve the problem step by step, we will analyze the motion of the car in three phases: acceleration, uniform motion, and deceleration.

### Step 1: Identify the phases of motion
The car goes through three phases:
1. Acceleration phase (from rest)
2. Uniform motion phase
3. Deceleration phase (to rest)

### Step 2: Define variables
- Acceleration \( a = 5 \, \text{m/s}^2 \)
- Total time of motion \( \tau = 25 \, \text{s} \)
- Average velocity \( \langle v \rangle = 72 \, \text{km/hr} = 20 \, \text{m/s} \)
- Let \( t \) be the time spent in the acceleration phase and the deceleration phase. Therefore, the time spent in uniform motion is \( 25 - 2t \).

### Step 3: Calculate the distance traveled during acceleration
Using the formula for distance during acceleration:
\[ s_1 = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ s_1 = \frac{1}{2} \cdot 5 \cdot t^2 = \frac{5}{2} t^2 \]

### Step 4: Calculate the velocity at the end of the acceleration phase
Using the formula for final velocity:
\[ v = u + at = 0 + 5t = 5t \]

### Step 5: Calculate the distance traveled during uniform motion
The distance during uniform motion is given by:
\[ s_2 = v \cdot t = (5t)(25 - 2t) \]
So,
\[ s_2 = 5t(25 - 2t) = 125t - 10t^2 \]

### Step 6: Calculate the distance traveled during deceleration
The distance during deceleration is the same as during acceleration:
\[ s_3 = \frac{1}{2} a t^2 = \frac{5}{2} t^2 \]

### Step 7: Total distance traveled
The total distance \( s \) is the sum of distances from all three phases:
\[ s = s_1 + s_2 + s_3 = \frac{5}{2} t^2 + (125t - 10t^2) + \frac{5}{2} t^2 \]
Combining the terms:
\[ s = 125t - 10t^2 + 5t^2 = 125t - 5t^2 \]

### Step 8: Use average velocity to find total distance
The average velocity is given by:
\[ \langle v \rangle = \frac{s}{\tau} \]
Substituting the values:
\[ 20 = \frac{125t - 5t^2}{25} \]
Multiplying both sides by 25:
\[ 500 = 125t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 125t + 500 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 25t + 100 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1} \]
\[ t = \frac{25 \pm \sqrt{625 - 400}}{2} \]
\[ t = \frac{25 \pm \sqrt{225}}{2} \]
\[ t = \frac{25 \pm 15}{2} \]
This gives us two possible values for \( t \):
1. \( t = \frac{40}{2} = 20 \)
2. \( t = \frac{10}{2} = 5 \)

### Step 10: Determine the valid value for \( t \)
Since the total time is 25 seconds, we cannot take \( t = 20 \) as it would imply the uniform motion time is negative. Thus, we take:
\[ t = 5 \, \text{s} \]

### Step 11: Calculate the time spent in uniform motion
The time spent in uniform motion is:
\[ \text{Time in uniform motion} = 25 - 2t = 25 - 2(5) = 25 - 10 = 15 \, \text{s} \]

### Final Answer
The car moves uniformly for **15 seconds**.

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