particle is projected vertically upwards with velocity $40m{s}^{-1}.$ Find the displacement and distance travelled by the particle in 6 s. $[Takeg=10m{s}^{-2}]$

Here, u is positive (upwards) and a is negative). So, first we will find $t_0,$ the time when velocity becomes zero i.e., when the particle is the highest point.

$t_0=\left|\frac{u}{a}\right|=\frac{40}{10}=4s$

Here, $t>t_0$

Hence,      distance > displacement

$s=40\times 6-\frac{1}{2}\times 10\times 36=60m$

While, $d=\left|\frac{u^2}{2a}\right|+\frac{1}{2}\left|a{(t-t_0)}^2\right|$

$=\frac{{\left(40\right)}^2}{2\times 10}+\frac{1}{2}\times 10\times {(6-4)}^2=100m$