A body travels 2m in the 2nd second and 6m in the next 4 seconds. What will be the distance travelled in 9th second.

Let's consider initial velocity is u and acceleration is a and 8th sec velocity is v..

we will this equation S=ut+0.5at² as shown below

2 = u*2 +0.5 *a*2*2 -----1
8 = u*6 + 0.5*a*6*6 ------2

now 3x(2 = u*2 +0.5 *a*2*2 ) ----3

now we will subtract from equation 3 to equation 2

we will get a = 1/6 m/s^2

Putting above value of a in eq 2 we get

8 = u*6+0.5*1/6*6*6

u = 5/6 m/s^2

v = u+at

Now, v = 5/6+ 1/6*9
So, v=7/3 m/s

Thanks & Regards,

Nirmal Singh

Askiitians Faculty