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Two particles P and Q are initially 40m apart P behind Q. P starts moving with uniform velocity of 10m/s towards Q. Q starts from rest and has acceleration 2m/s2 in the dorection of P. Then minimum distance between P and Q will be ?
Two particles P and Q are initially 40m apart P behind Q. P starts moving with uniform velocity of 10m/s towards Q. Q starts from rest and has acceleration 2m/s2  in the dorection of P. Then minimum distance between P and  Q will be ?

```
4 years ago

## Answers : (1)

Vikas TU
14149 Points
```							From first law of eqn.v = u + at10= 0 + 2*tt = 5s.In this timeP would have moved 5*10 = 50 m  Q moves 5*5 = 25 m.THERFEORE, minimum distance  it would cover = >40 + 25 – 50 = 15 m.
```
4 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions