43. A particle travels 10 m in first 5 sec and 10 m in next
3 sec. Assuming constant acceleration what is the
distance travelled in next 2 sec
(a) 8.3 m
(b) 9.3 m
(c) 10.3 m
(d) None of above
Answers
Answer:
Explanation:
Given :-
Time taken by particle, t = 5 seconds
Distance covered by particle, S = 10m/s.
Using the second equation of motion,
Distance covered in first 5 sec-
10= 5u + 1/2(a × 25) ------ (1)
Given :-
Time taken by particle, t = 5 + 3 = 8 seconds
Distance covered by particle, S = 20 m.
Using the second equation of motion,
Distance covered in next 3 sec-
10 = 3u + 1/2(a × 9) --------(2)
To Find :-
Distance traveled.
Solution :-
On solving these 2 equations, we get value
Velocity, u = 16/3
Acceleration a = -1.33
So the distance covered by particle in last 2 sec is-
S = 16/3 × 2 - 1/2 × 1.33 × 4
= 8 m
Hence, the distance traveled in next 2 seconds is 8 meter.
Answer:
8.25 m
Explanation:
•particles move=10 m and t=5 s
finding distance by using 2nd equation
S=ut+1/2at^2
10=u×5+1/2×a×5×5
10=5u+1/2×25a
10=5u+25a .........i)
•Again particles move=10 m,t=8 s and S=20
S=ut+1/2at^2
20=u×8+1/2×a×8×8
20=8u+64a/2 .......ii)
Solving i) and ii)
a=1/3 and u=7/6
Now for next 2 sec t=10
S=7/6×10+1/2×1/3×10×10
=170/6
Distance travelled in 2 sec=170/6-20
=50/6
=8.25 m