Car A and car B start moving simultaneously in the same direction along the line joining them. Car A moves with a constant acceleration `a=4 m//s^2,` while car B moves with a constant velocity `v=1 m//s.` At time `t=0,` car A is `10 m` behind car B. Find the time when car A overtake car B.

Explanation

To solve the problem of when car A overtakes car B, we can follow these steps:

### Step 1: Define the initial conditions
- Car A starts 10 meters behind car B.
- Car A has an initial velocity \( u_A = 0 \) m/s and accelerates at \( a_A = 4 \) m/s².
- Car B moves with a constant velocity \( v_B = 1 \) m/s and has an initial velocity \( u_B = 1 \) m/s.

### Step 2: Write the equations of motion for both cars
1. **For Car A** (which has acceleration):
\[ S_A = u_A t + \frac{1}{2} a_A t^2 \]
Substituting the values:
\[ S_A = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]

2. **For Car B** (which has constant velocity):
\[ S_B = u_B t + d_0 \]
Here, \( d_0 = 10 \) m (the initial distance ahead of car A):
\[ S_B = 1 \cdot t + 10 = t + 10 \]

### Step 3: Set the distances equal to find when car A overtakes car B
To find when car A overtakes car B, we set \( S_A = S_B \):
\[ 2t^2 = t + 10 \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ 2t^2 - t - 10 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 2 \), \( b = -1 \), and \( c = -10 \).
- Calculate the discriminant:
\[ D = b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot (-10) = 1 + 80 = 81 \]
- Now, apply the quadratic formula:
\[ t = \frac{-(-1) \pm \sqrt{81}}{2 \cdot 2} = \frac{1 \pm 9}{4} \]
This gives us two potential solutions:
1. \( t = \frac{10}{4} = 2.5 \) seconds
2. \( t = \frac{-8}{4} = -2 \) seconds (not physically meaningful)

### Step 6: Conclusion
Thus, car A overtakes car B after **2.5 seconds**.

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