A bus starts from rest with an acceleration of 1m/s . A man who is 48m behind the bus starts with a uniform velocity 10 m/s . Find minimum time after which the man will catch the bus.
Answers
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Answer:
- Minimum time after which Man will catch the Bus = 8 seconds
Explanation:
Given :
- Bus starts from rest, u = 0
- Acceleration of Bus, a = 1 m/s²
- Initial distance of Man from Bus, d = 48 m
- Man starts moving with a uniform velocity, V = 10 m/s
To find :
- Minimum time after which Man will catch the bus, t =?
Solution :
Let, Man catches the Bus after travelling for a time t, such that Bus travels a distance of x metres till then.
then, distance covered by Man before catching the Bus will be ( x + 48 ) metres.
Now,
For Bus Using second equation of motion :
- s = u t + 1/2 a t²
[ Where s is distance covered, u is initial velocity, t is time taken, a is acceleration ]
→ x = u t + 1/2 a t²
→ x = ( 0 ) t + 1/2 ( 1 ) ( t² )
→ x = t² / 2 _____equation (1)
Further,
For man Using formula :
- v = s / t
[ where s is displacement, t is time taken and v is uniform velocity ]
→ 10 = ( x + 48 ) / t
→ x + 48 = 10 t
Using equation (1)
→ (t²/2) + 48 = 10 t
→ t² + 96 = 20 t
→ t² - 20 t + 96 = 0
→ t² - 12 t - 8 t + 96 = 0
→ t ( t - 12 ) - 8 ( t - 12 ) = 0
→ ( t - 8 ) ( t - 12 ) = 0
so,
→ t = 8 sec or t = 12 sec
It means Man could catch the Bus after 8 sec or 12 sec.
Since, we needed to find Minimum value, therefore
- Minimum time after which Man will catch the Bus will be 8 seconds.