A particle accelerates from rest with an acceleration of 5 m/s2. After sometime, particle moves with constant velocity. Then it decelerates with 5 m/s2 and finally comes to rest. In whole of the journey it takes 25 sec and the average speed in the journey is 72 km/hr. Then which are correct?
A) The time interval during which it has moved with constant velocity is 15 sec.
B) The max velocity acquired by it is 25 m/s
C) The time interval during which it has moved with constant velocity is 25 sec.
D) The max velocity acquired by it is 30 m/s
Answers
Welcome Dear,
◆ Answer -
- A) The time interval during which it has moved with constant velocity is 15 sec.
- C) The time interval during which it has moved with constant velocity is 25 sec.
● Explanation -
During acceleration,
s = ut + 1/2 at^2
s1 = (5/2) t1^2
During constant velocity -
s = v × t
s2 = 5 t1.t2
During deceleration,
s = ut + 1/2 at^2
s3 = 5 t1.t3 - (5/2) t3^2
Average speed given is -
v = 72 km/hr = 20 m/s
Hence, total distance can be calculated by -
s = v × t
s = 20 × 25
s = 500 m
Hence, now
s = s1 + s2 + s3
500 = (5/2) t1^2 + 5 t1.t2 + 5 t1.t3 - (5/2) t3^2
But same time will be taken for acceleration & deceleration. i.e. t3 = t1.
Now solving
t1 = 5 s
t2 = 15 s
t3 = 5 s
s1 = 62.5 m
s2 = 375 m
s3 = 62.5 m
vmax = 25 m/s
Therefore, maximum velocity attained is 25 m/s and time interval with constant velocity is 15 s.
Hope this helped you.
Answer:
Welcome Dear,
◆ Answer -
A) The time interval during which it has moved with constant velocity is 15 sec.
C) The time interval during which it has moved with constant velocity is 25 sec.
● Explanation -
During acceleration,
s = ut + 1/2 at^2
s1 = (5/2) t1^2
During constant velocity -
s = v × t
s2 = 5 t1.t2
During deceleration,
s = ut + 1/2 at^2
s3 = 5 t1.t3 - (5/2) t3^2
Average speed given is -
v = 72 km/hr = 20 m/s
Hence, total distance can be calculated by -
s = v × t
s = 20 × 25
s = 500 m
Hence, now
s = s1 + s2 + s3
500 = (5/2) t1^2 + 5 t1.t2 + 5 t1.t3 - (5/2) t3^2
But same time will be taken for acceleration & deceleration. i.e. t3 = t1.
Now solving
t1 = 5 s
t2 = 15 s
t3 = 5 s
s1 = 62.5 m
s2 = 375 m
s3 = 62.5 m
vmax = 25 m/s
Therefore, maximum velocity attained is 25 m/s and time interval with constant velocity is 15 s.
Hope this helped you.
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Explanation: