Question

A particle moves 10 m in 4th second and 15 m in 6th second of its motion.calculate the distance travelled by it in 7th second if the motion of the particle is uniformly acelerated

Answer 1

Given :-

Distance travelled in 4th sec =${S}_{4}$ = 10 m

Distance travelled in 6th sec =${S}_{6}$ = 15 m

${S}_{4}$ = u + a/2[2×4-1]

10 = u + 7a/2 -----(1)

Again,

${S}_{6}$ = u + a/2[2×6-1]

15 = u + 11a/2 ----(2)

(2)-(1)

u + 11a/2 - u - 7a/2 = 15 - 10

4a/2 = 5

a = 5/2 ms-²

Putting the value of 'a' in eq(1) to get 'u'.

10 = u + (5/4)[7]

10 = u + 35/4

u = 10 - 35/4

u = 40-35/4

u = 5/4 ms-¹

Again,

${S}_{7}$ = 5/4 + 5/4[2×7-1]

${S}_{7}$ = 5/4 + 65/4

${S}_{7}$ = 70/4 m

$\bf{S}_{7} = 17.5\:m$

$\bf{Answer}$