Two cars `A` and `B` cross a point `P` with velocities `10m//s` and `15m//s`. After that they move with different uniform accelerations and the car `A` overtakes `B` with a speed of `25ms^(-1)`. What is velocity of `B` at that instant?

Options

A. (a)`20ms^(-1)`
B. (b)`25ms^(-1)`
C. (c)`30ms^(-1)`
D. (d)`40ms^(-1)`

Answer

To solve the problem, we need to analyze the motion of both cars A and B.

### Step-by-Step Solution:

1. **Identify Initial Velocities**:
- Car A crosses point P with an initial velocity \( u_A = 10 \, \text{m/s} \).
- Car B crosses point P with an initial velocity \( u_B = 15 \, \text{m/s} \).

2. **Final Velocity of Car A**:
- At the point where car A overtakes car B, its velocity is given as \( v_A = 25 \, \text{m/s} \).

3. **Use the Equation of Motion**:
- We will use the equation of motion \( v = u + at \) for both cars.
- For car A:
\[ v_A = u_A + a_A t \implies 25 = 10 + a_A t \implies a_A t = 15 \implies a_A t = 15 \]
- For car B:
\[ v_B = u_B + a_B t \implies v_B = 15 + a_B t \]

4. **Distance Covered by Both Cars**:
- Since both cars cover the same distance when A overtakes B, we can set their distance equations equal to each other.
- Distance covered by A:
\[ d = u_A t + \frac{1}{2} a_A t^2 = 10t + \frac{1}{2} a_A t^2 \]
- Distance covered by B:
\[ d = u_B t + \frac{1}{2} a_B t^2 = 15t + \frac{1}{2} a_B t^2 \]

5. **Equate the Distances**:
- Setting the distances equal gives:
\[ 10t + \frac{1}{2} a_A t^2 = 15t + \frac{1}{2} a_B t^2 \]
- Rearranging gives:
\[ \frac{1}{2} a_A t^2 - \frac{1}{2} a_B t^2 = 5t \]
- Factoring out \( t \) (assuming \( t \neq 0 \)):
\[ \frac{1}{2} (a_A - a_B) t = 5 \implies a_A - a_B = \frac{10}{t} \]

6. **Substituting \( a_A t \)**:
- From step 3, we have \( a_A t = 15 \), so:
\[ a_A = \frac{15}{t} \]
- Substitute into the equation:
\[ \frac{15}{t} - a_B = \frac{10}{t} \implies a_B = \frac{15}{t} - \frac{10}{t} = \frac{5}{t} \]

7. **Finding Velocity of Car B**:
- Now substitute \( a_B \) back into the equation for \( v_B \):
\[ v_B = 15 + a_B t = 15 + 5 = 20 \, \text{m/s} \]

### Final Answer:
The velocity of car B at the instant when car A overtakes it is \( \boxed{20 \, \text{m/s}} \).