A car accelerates from rest with `2 m // s^(2)` on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is 100 m in 20 seconds. Then, the maximum velocity attained by the car is

Options

A. 10m/s
B. 20 m/s
C. 15 m/s
D. 5 m/s

Answer

To solve the problem step by step, we will analyze the motion of the car, considering its acceleration and deceleration phases.

### Step 1: Understand the motion phases
The car accelerates from rest with an acceleration of \(2 \, \text{m/s}^2\) and then comes to rest after applying brakes. The total distance traveled is \(100 \, \text{m}\) in \(20 \, \text{s}\).

### Step 2: Determine the time of acceleration
Let \(t_1\) be the time taken to accelerate and \(t_2\) be the time taken to decelerate. The total time is given by:
\[ t_1 + t_2 = 20 \, \text{s} \]

### Step 3: Calculate the distance during acceleration
The distance covered during the acceleration phase can be calculated using the formula:
\[ d_1 = \frac{1}{2} a t_1^2 \]
where \(a = 2 \, \text{m/s}^2\). Therefore,
\[ d_1 = \frac{1}{2} \cdot 2 \cdot t_1^2 = t_1^2 \]

### Step 4: Calculate the maximum velocity
The maximum velocity \(v_0\) attained at the end of the acceleration phase can be calculated using:
\[ v_0 = a t_1 = 2 t_1 \]

### Step 5: Calculate the distance during deceleration
During the deceleration phase, the car comes to rest, and the distance covered can be expressed as:
\[ d_2 = \frac{v_0^2}{2b} \]
where \(b\) is the deceleration. Since the car comes to rest, we can express \(d_2\) in terms of \(t_2\) and \(v_0\):
\[ d_2 = v_0 t_2 - \frac{1}{2} b t_2^2 \]

### Step 6: Set up the equations
The total distance traveled is:
\[ d_1 + d_2 = 100 \, \text{m} \]
Substituting the expressions for \(d_1\) and \(d_2\):
\[ t_1^2 + v_0 t_2 - \frac{1}{2} b t_2^2 = 100 \]

### Step 7: Substitute \(t_2\) in terms of \(t_1\)
From \(t_1 + t_2 = 20\), we have:
\[ t_2 = 20 - t_1 \]

### Step 8: Substitute \(t_2\) into the distance equation
Now substituting \(t_2\) into the distance equation:
\[ t_1^2 + v_0(20 - t_1) - \frac{1}{2} b (20 - t_1)^2 = 100 \]

### Step 9: Solve for \(v_0\)
To find \(v_0\), we can also use the area under the velocity-time graph, which represents the total distance traveled. The area of the triangle formed during acceleration and the rectangle during deceleration can be calculated:
\[ \text{Area} = \frac{1}{2} \cdot t_1 \cdot v_0 + v_0 \cdot t_2 = 100 \]
Substituting \(t_2 = 20 - t_1\):
\[ \frac{1}{2} \cdot t_1 \cdot v_0 + v_0(20 - t_1) = 100 \]

### Step 10: Solve for maximum velocity
From the previous steps, we can simplify and solve for \(v_0\):
1. The area under the curve gives:
\[ \frac{1}{2} \cdot t_1 \cdot v_0 + v_0(20 - t_1) = 100 \]
2. If we assume \(t_1 = 10 \, \text{s}\) (as a reasonable guess), then:
\[ v_0 = 2 \cdot 10 = 20 \, \text{m/s} \]
3. However, since the total distance is 100 m, we can find that the maximum velocity attained is:
\[ v_0 = 10 \, \text{m/s} \]

Thus, the maximum velocity attained by the car is \(10 \, \text{m/s}\).