Question

A particle, moving with a uniform Acceleration describes a distance of
720 cm in 11th see and a distance of 960 cm in 15th sec. Find the
initial velocity and acceleration of the particle.​

Answer 1

Answer:

displacement for the n

th

second is given by ,

S

n

=u+

2

1

a(2n−1)

so from the question we get ,

S

3

=u+

2

1

a(2×3−1)⇒12=u+(5/2)a....(1)

S

5

=u+

2

1

a(2×5−1)⇒20=u+(9/2)a...(2)

subtracting eq1 from eq 2 we get

8=2a⇒a=4

usin this in eq 1 we get value of u=2

now for solving for the eighth second ,

S

8

=2+

2

1

×4×(2×8−1)=32m