A particle moving in a straight line with uniform acceleration is observed to be a distance a from a fixed point initially.It is at distance b,c,d from the same point after n,2n,3n seconds.The acceleration of the particle is
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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
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