A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec?
Answers
25/3 = 8.33 m distance travelled in next 2 sec
Explanation:
Let say initial velocity = u m/s
acceleration = a m/s^2
using S = ut + (1/2)at^2
10 = 5u + 25a/2
20 = 10u + 25a
4 = 2u + 5a
in next 3 secs 10 m
hence 5 +3 = 8 secs foe 10+10 = 20 m
20 = 8u + 32a
5 = 2u + 8a
from both 3a = 1
a = 1/3
4 = 2u + 5/3
2u = 7/3
u = 7/6
distance covered in next 2 sec
= in 10 secs - 20 m
10u + 50a - 20
= 70/6 + 50/3 - 20
= 25/3
= 8.33 m
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Let say initial velocity = u m/s
acceleration = a m/s^2
using S = ut + (1/2)at^2
10 = 5u + 25a/2
20 = 10u + 25a
4 = 2u + 5a
in next 3 secs 10 m
hence 5 +3 = 8 secs foe 10+10 = 20 m
20 = 8u + 32a
5 = 2u + 8a
from both 3a = 1
a = 1/3
4 = 2u + 5/3
2u = 7/3
u = 7/6
distance covered in next 2 sec
= in 10 secs - 20 m
10u + 50a - 20
= 70/6 + 50/3 - 20
= 25/3
= 8.33 m