A bus starts moving with acceleration `2 ms^-2`. A cyclist `96 m` behind the bus starts simultaneously towards the bus at a constant speed of `20 m//s`. After what time will he be able to overtake the bus ?

Options

A. 4 s
B. 8 s
C. 12 s
D. 16 s

Explanation

To solve the problem of when the cyclist will overtake the bus, we can break it down into a few steps.

### Step 1: Understand the motion of the bus
The bus starts from rest and accelerates at a rate of \(2 \, \text{m/s}^2\). We can use the equation of motion to find the distance covered by the bus after time \(t\):
\[ x_{\text{bus}} = u_{\text{bus}} t + \frac{1}{2} a_{\text{bus}} t^2 \]
Since the initial speed \(u_{\text{bus}} = 0\), the equation simplifies to:
\[ x_{\text{bus}} = \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \]

### Step 2: Understand the motion of the cyclist
The cyclist starts \(96 \, \text{m}\) behind the bus and moves towards it at a constant speed of \(20 \, \text{m/s}\). The distance covered by the cyclist after time \(t\) is:
\[ x_{\text{cyclist}} = 20t \]
However, since the cyclist starts \(96 \, \text{m}\) behind the bus, the effective distance he needs to cover to reach the bus is:
\[ x_{\text{cyclist}} = 96 + x_{\text{bus}} = 96 + t^2 \]

### Step 3: Set the distances equal
To find the time when the cyclist overtakes the bus, we set the distances equal:
\[ 20t = 96 + t^2 \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ t^2 - 20t + 96 = 0 \]

### Step 5: Solve the quadratic equation
Now we can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -20\), and \(c = 96\):
\[ t = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 96}}{2 \cdot 1} \]
\[ t = \frac{20 \pm \sqrt{400 - 384}}{2} \]
\[ t = \frac{20 \pm \sqrt{16}}{2} \]
\[ t = \frac{20 \pm 4}{2} \]
This gives us two possible solutions:
\[ t = \frac{24}{2} = 12 \quad \text{and} \quad t = \frac{16}{2} = 8 \]

### Step 6: Determine the valid time
Since the question asks for the time after which the cyclist will overtake the bus, we consider the first instance:
\[ t = 8 \, \text{seconds} \]

Thus, the cyclist will overtake the bus after **8 seconds**.