MHT CET 202623 AprilMorning Shift
- Exam
- MHT CET
- Year
- 2026
- Date
- 23 April
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Motion in One Dimension
MHT CET 2026 · 23 April · Morning Shift · Physics · Motion in One Dimension
A train accelerates from rest at a constant rate 'p' for some time...
A train accelerates from rest at a constant rate 'p' for some time (attains velocity 'v') and then retards to rest at the constant rate 'q'. If the total distance covered by the train is 's', then the velocity 'v' of the train is
Options
- A\left( \left( \dfrac{p+q}{2pq} \right) \times s \right)^{\frac{1}{2}}
- B\left( \left( \dfrac{p-q}{2pq} \right) \times s \right)^{\frac{1}{2}}
- C\left( \left( \dfrac{2pq}{p+q} \right) \times s \right)^{\frac{1}{2}}
- D\left( \left( \dfrac{2pq}{p-q} \right) \times s \right)^{\frac{1}{2}}
Correct answer
C. \left( \left( \dfrac{2pq}{p+q} \right) \times s \right)^{\frac{1}{2}}
Step-by-step solution
Let be the distance covered during acceleration and be the distance covered during retardation.
Using the third equation of motion for the acceleration phase:
For the retardation phase:
Total distance
Solving for :
v = \left(\left( \dfrac{2pq}{p+q} \right) \times s \right)^{\dfrac{1}{2}}
Using the third equation of motion for the acceleration phase:
For the retardation phase:
Total distance
Solving for :
v = \left(\left( \dfrac{2pq}{p+q} \right) \times s \right)^{\dfrac{1}{2}}