Question

for an object moving with a uniform acceleration and travelling 50 in 5 th second and 70m in 7 th s its initial velocity and acceleration and average velocity during 9, th second is
please answer I will mark as branliest​

Answer 1

Answer:

With the 4th equation of motion =

S(n) = u + a/2 × ( 2n-1)

n in the first case = 5

S(n) in the first case = 50m

therefore, => 50 = u + a/2 × (9) --->(equation 1)

S(n) in the second case = 70m

n here = 7

therefore, => 70 = u + a/2 × ( 2×7 - 1)

=> 70 = u + a/2 × (13) ---> ( equation 2)

subtracting equation 1 from 2 :-

20 = (a/2) × (4)

=> a/2 = 5

=> a = 10

therefore, acceleration = 10m/s²

by putting the value of a in equation 1 :-

=> 50 = u + 10/2 × (9)

=> 50 = u + 45

=> u = 5m/s

by putting the values of a and u in the third equation :-

S(n) = u + a/2 × ( 2×9 - 1)

=> S(n) = 5+ 10/2 ( 17)

=> S(n) = 5+85

=> S(n) = 90m

So , average velocity of the given values = (V1+V2+V3)/3

Using first equation of motion :-

velocity in 5th sec => u + at

= 5+50

= 55m/s

in 7th sec :-

= u+at

= 5+70

= 75m/s

in 9th sec :-

= u+at

= 5+90

= 95m/s

average velocity in 9th sec = all velocities provided / number of velocities

= 55+75+95/3

= 225/3

= 75m/s

Hence , 75m/s is the average velocity of the whole journey of 9s.

Hope it helped :)