Question

Initially a body is at rest. if its acceleration is 5ms^-2 then the distance travelled in the 18th second is:

Answer 1

At the end of 17 s its speed is 0 + 5 *17 = 85.

After this in the next second the speed increases by 5 m/s.

Average speed in the 18 th second is 85 + (5/2) = 87.5 m/s.

Distance traveled is 87.5*1 = 87.5 m.

Answer 2

Given that,

Initial speed, u = 0

Acceleration, a = 5 m/s²

To find,

The distance traveled in the 18th second.

Solution,

We have a formula that gives the distance traveled by the body in nth seconds. It is as follows :

$S_n=u+\dfrac{1}{2}a(2n-1)$

Put n = 18

$S_{18}=0+\dfrac{1}{2}\times 5\times (2(18)-1)\\\\S_{18}=87.5\ m$

So, the distance traveled in the 18th second is 87.5 m.