a particle starts moving from rest in a straight line, first with acceleration a=10m/s² then with uniform motion and finally deaccelerating at same rate 'a' and comes to rest. if average velocity of journey during total time of 160s is 300m/s, then time for which the particle moves uniformly is --
Answers
Answer:
80 seconds
Explanation:
Let, the length of
- 1st part = x
- 2nd part = y
- 3rd part = z
Time taken in
- 1st part = t
- 2nd part = t'
- 3rd part = t''
In first part it's accelerating starting from rest. So, from equation of motion,
- x = 0.5at² = (0.5 × 10 × t²) = 5t²
In second part it's in uniform motion with velocity "v". This velocity is equal to the velocity at the end of 1st part. It can be calculated from equation of motion as,
- v = u + at = 0 + (10t) = 10t
Then,
- y = vt' = (10t)t'
In third part it's decelerating. It starts with velocity "v" and stops after traveling for time (t'')
So, from equation of motion
- z = (0² - v²)/(-2a)
- z = v²/(2a)
- z = (10t)²/(2 × 10)
- z = 5t²
Time taken in third part from equation of motion is
- t'' = (0 - v)/(-10) = v/10
- t'' = 10t/10
- t'' = t
Total time given is 160 s
So,
- t + t' + t'' = 160
- t + t' + t = 160
- 2t + t' = 160
- t' = 160 - 2t
Average velocity = Total displacement / Total time
300 m/s = (x + y + z) / (160 s)
x + y + z = 48000 m
5t² + (10t)t' + 5t² = 48000
10t² + (10t)t' = 48000
t² + tt' = 4800
t² + t(160 - 2t) = 4800
t² + 160t - 2t² = 4800
-t² + 160t - 4800 = 0
t² - 160t + 4800 = 0
t² - 120t - 40t + 4800 = 0
(t - 120)(t - 40) = 0
t = 120 (or) t = 40
If t = 120 s then
t = t'' = 120
t + t' + t'' = 160
120 + t' + 120 = 160
t' = -80 s
This is not possible since time can't be negative
If t = 40 s then
t = t'' = 40 s
t + t' + t'' = 160
40 + t' + 40 = 160
t' = 80 seconds