Institution: বঙ্গবন্ধু শেখ মুজিবুর রহমান বিজ্ঞান ও প্রযুক্তি বিশ্ববিদ্যালয় (BSMRSTU) Exam: A ইউনিট || বঙ্গবন্ধু শেখ মুজিবুর রহমান বিজ্ঞান ও প্রযুক্তি বিশ্ববিদ্যালয় || 2011
Topic: Physics (পদার্থবিদ্যা), Subject: General Science (সাধারণ বিজ্ঞান)

Best Answer Explained in Below-:

To solve the problem, we start by using the kinematic equation for distance traveled under constant acceleration:

\[ s = ut + \frac{1}{2}at^2 \]

where \( s \) is the distance, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

For Car A:
- Initial speed, \( u_A = 10 \, \text{m/s} \)
- Acceleration, \( a_A = 2 \, \text{m/s}^2 \)

Using the equation, the distance covered by Car A after time \( t \) is:
\[ s_A = 10t + \frac{1}{2}(2)t^2 = 10t + t^2 \]

For Car B:
- Initial speed, \( u_B = 5 \, \text{m/s} \)
- Acceleration, \( a_B = 3 \, \text{m/s}^2 \)

The distance covered by Car B is:
\[ s_B = 5t + \frac{1}{2}(3)t^2 = 5t + \frac{3}{2}t^2 \]

Since both cars reach the destination at the same time, we set \( s_A = s_B \):
\[ 10t + t^2 = 5t + \frac{3}{2}t^2 \]

Rearranging this gives:
\[ t^2 - \frac{1}{2}t = 0 \]
Factoring out \( t \):
\[ t(t - \frac{1}{2}) = 0 \]
Ignoring \( t=0 \), we find:
\[ t = 10 \, \text{s} \]

Thus, the duration or the time taken by both cars to reach their destination is 10 seconds, confirming option **C) 10 s** as the correct answer.