A particle having initial velocity u moves with a constant acceleration a for a time t. a. Find the displacement of the particle in the last 1 second . b. Evaluate it for `u=5m//s, a=2m//s^2 and t=10s`.

Explanation

a. The positon at time t is
`s=ut+1/2 at^2`
The position at time `(t-1s)` is ltbr.`s'=u(t-1s)+1/2a(t-1s)^2`
`ut-u(1s)+1/2at^2-at(1s)=1/2a(1s)^2`
Thus, the displacement in the last 1 s is
`s_t=s-s'`
`=u(1s)+at)1s)-1/2a(1s)^2`
or, ` s_t=u(1s)+a/2(2t-1s)(1s).` ......i
b. Putting the givne vaues in i.
`s_t=(5m/s0(1s)=1/2(2 m/s^2)(2xx10-s-1s)(1s)`
`=5m+(1 m/s^2)(19s)(1s)`
` 5m+19m=24m`
sometimes, we are not creful in writing the units appearing withteh numerical values of physical quantities. If we forget to write the unit of second in equation i. we get,
`s_t=u+a/2(2t-1)`.
This equation is often used to calculate the displacement in the t th second. However, as you can verify, different terms in the equation have different dimensions and hence the above equation is dimensioN/Ally incorrect. Equation i. is the correct form which was used to solve part b.
Also note that this equation gives the displacement of the particle in tehlast 1 second and not necessarily the distance covered in that second.