A particle moves with uniform acceleration along a straight line. it covers 41 cm and 49 cm in the 6th and 10th sec. what will be the distance covered by the particle in 15th sec.
Answers
Answer:
The distance travelled during the nth second of motion of a body is given by
Sñ = u + 1/2 a(2n-1)
for the motion of during the 8th second,
3 = u + 1/2 a(16 - 1) = u + 15a/2 ..........(I)
for the motion during the 16th second,
5 = u + 1/2 a(32 - 1) = u + 31a/2 ...........(ii)
Now,
subtracting equation (I) from (ii)
8a = 2
acceleration (a) = 1/4 m/sec²
from equation (I) , u = 3 - (15/2 * 1/4 ) = 9/8 m/sec
Now,
the velocity at the end of 5 second
V1 = u + 5a
average velocity during this interval of 10 seconds
= v1 + v2/2
= (u + 5a) + (u + 15a)/2
= u + 10 a
distance travel during the interval
S = average velocity * time
= ( u + 10a ) * t
= (9/8 + 10/4) * 10
= 290/8
= 36.25m...
hope it helps
Explanation:
The distance travelled in n
th
second is
S
n
=u+
2
1
(2n−1)a ....(1)
So distance travelled in t
th
&(t+1)
th
second are
S
t
=u+
2
1
(2t−1)a ....(2)
S
t+1
=u+
2
1
(2t+1)a ....(3)
As per question,
S
t
+S
t+1
=100=2(u+at) ....(4)
Now from first equation of motion the velocity of particle after time t, if it moves with an acceleration a is
v=u+at ....(5)
where u is initial velocity
So from equation (4) and (5), we get v=50cm/s
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