Question

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

Answer 1

Answer:

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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