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 follow these steps:

### Step 1: Understand the problem
The bus starts moving with an acceleration of \(2 \, \text{m/s}^2\) from rest, while the cyclist starts \(96 \, \text{m}\) behind the bus and moves towards it at a constant speed of \(20 \, \text{m/s}\).

### Step 2: Set up the equations of motion
1. **Equation for the bus**: The distance traveled by the bus can be described by the second equation of motion:
\[ s_b = ut + \frac{1}{2} a t^2 \]
where:
- \(u = 0\) (initial velocity of the bus)
- \(a = 2 \, \text{m/s}^2\)
- \(s_b\) is the distance traveled by the bus in time \(t\).

Thus, the equation simplifies to:
\[ s_b = 0 + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \]

2. **Equation for the cyclist**: The distance traveled by the cyclist is given by:
\[ s_c = vt \]
where:
- \(v = 20 \, \text{m/s}\)
- The cyclist starts \(96 \, \text{m}\) behind the bus, so the effective distance he needs to cover is \(s_c = 20t\).

### Step 3: Set up the equation for overtaking
To find the time when the cyclist overtakes the bus, we need to set the distance traveled by the bus equal to the distance traveled by the cyclist plus the initial distance between them:
\[ s_b = s_c + 96 \]
Substituting the equations we derived:
\[ t^2 = 20t + 96 \]

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

### Step 5: Solve the quadratic equation
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 \(c = -96\).
Calculating the discriminant:
\[ b^2 - 4ac = (-20)^2 - 4 \cdot 1 \cdot (-96) = 400 + 384 = 784 \]
Now substituting into the quadratic formula:
\[ t = \frac{20 \pm \sqrt{784}}{2 \cdot 1} = \frac{20 \pm 28}{2} \]
This gives us two possible solutions:
\[ t_1 = \frac{48}{2} = 24 \quad \text{and} \quad t_2 = \frac{-8}{2} = -4 \]
Since time cannot be negative, we take \(t = 24 \, \text{s}\).

### Step 6: Conclusion
The cyclist will overtake the bus after \(24\) seconds.

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