Question

a body sliding on a smooth inclined plane requires 9 seconds to reach the bottom starting from rest at the top the distance it to will take to cover one-fourth distance starting from rest at the top is

Answer 1

Explanation:

s=ut+(1÷2)at^2

here u = 0

a = 10 m/s for easier calculation

for t = 9

s = 405m

now for 405÷4 distance

405÷4 = 1÷2 × 10 × t^2

solving this we get t = 6.36

Answer 2

Explanation:

Time to reach the bottom = 9 seconds

let

a = acceleration

S = distance to the bottom

t = time

S = 1/2 × a × t²

S = 1/2×a×9²

S = 1/2×a×81

S = 81a/2 ------ equation 1

Let the time of sliding of S/4 from the top Starting from rest be t seconds

S/4 = 0 × t + 1/2 × a × t²

S/4 = 1/2 ×at² ----- equation 2

comparing both equations

81 a/2 = 1/2×at²

81 = t²

t = √81

t = 9 seconds