Question

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

Answer 1

Chotu here u go with your answer___

✔Let u, v and a be the initial velocity, final velocity and acceleration of the body respectively.

As we know,

Sn=u+a/2 * (2n-1) ………(1)

✔where Sn= distance in nth second and it is only one term here not S*n it should be remind.

S= ut+a/2 * t^2 ………….(2)

as we have to find the distance in 9th second so we will use the first formula(1) here.


✔As in question it is given that body covers 2 meter distance in 2nd second.

so by using first formula (1)

here n=2 & Sn=2 by putting values we get

2=u+a/2 * {2(2)-1}

2=u+a/2 *3

2=u+3a/2

multiply by 2 on both side

4=2u+3a …………..(3)

and for the second equation it is given that distance covered in next foursecond is 6 meter.

so we know that it (distance in next four seconds) is the distance that is covered by body in 2 to 6 seconds interval. and we can say this same statement that it is the distance that is the “ Difference of the distances covered in 6 seconds and 2 seconds.”

✔so in mathematical statement you can write it as:

S(2 to 4)=S6-S2

6= (S in 6 seconds) - (S in 1st 2 seconds)

by using the second formula (b) to find distances in 6 and first 2 seconds.

6=(ut+a/2 * t^2) - (ut+a/2 * t^2)

6=[{u(6)+a/2 * (6)^2} - {u(2)+a/2 * (2)^2}]

6={6u+a/2 * (36)} - {2u+a/2 * (4)}

6=(6u+18a) - (2u+2a)

6=6u+18a-2u-2a

6=4u+16a

dividing by 2 on both side we get

3=2u+8a …………..(4)

so now (3) and (4) are two simultaneous equation by solving, we get.

u=2.3 m/s
a= -0.2 m/s^2

so now use above formula (1) and calculate distance in 9th second.

Sn=u+a/2 * (2n-1)

here S9 means distance in 9th second.

S9=2.3+(-0.2)/2 *{2(9)-1}

S9=2.3-(0.2)/2 *{2(9)-1}

S9=2.3-(0.1)(17)

S9=(2.3–1.7)

✔S9= 0.6 Answer.

✔Hope this answer will help u Chotu..


@coco didi