Question

A particle moving with uniform acceleration in a straight line covers a distance of 3 m in the 8th second and 5 metre in the 16th second of its motion. what is the displacement of the particle from the beginning of 6th second to the end of 15th second?

Answer 1

ANSWER :-

$solution$

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

_______________________

❤BE BRAINLY ❤

Answer 2

Answer:

36.25 m

Explanation:

Let u be the initial velocity, a be the acceleration of the particle. Distance covered by the particle in 8th second is 3 m .using the equation for Snth ,

3 = u +1/2 a(2* 16-1) or 3= u + 1/2 a(15) or

2u+15a= 6------ 1 equation

distance covered by the particle in 16th second is 5m . Again ,using the equation for Snth,

5=u +1/2 a(2*16-1) or 2u+31a=10

Equation 2 -Equation 1

(2u+31a) - (2u+15a)= 10-6

or 16a = 4 or a = (1/4) m/s^2

Using equation 1 we get 2u+15 1 /4 = 6 or

2 u = 6 - 15/4=9/4 or u = (9/8) m/s

Now, we have to find the distance covered by the particle from the beginning of the 6th second to the end of the 15th second . At the beginning of the 6th second, total time elapsed in 5 second. First, we will find the velocity at the end of velocity the end of 5th second using first equation of motion,

v=u+at or v = 9/8+ (1/4) (5)= 9/8+ 5/4= 19/8 m/s

now time taken between the beginning of sixth second to the end of the 15th second is actually 10 seconds (6th, 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th). [Caution : if you subtract 15 - 6, you will get 9 seconds while actual time elapsed is 10 seconds]

Now, using second equation of motion,

s = ut + 1/2 at ^2 , we get ,

s = (19/8)(10)+ 1/2 (1/4) 10 ^ 2 = 190/8 + 100/8 = 290/8 = 36.25 m