Question

An object starts from rest with constant acceleration 4 m/s²,then find the distance travelled by object in 5^th half second.

Answer 1

$$\begin{gathered} \mathrm{Distance}\mathrm{travelled}\mathrm{in}2.5\mathrm{s}: \\ {\mathrm{s}}_1=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ =\left(0\right)\left(2.5\right)+\frac{1}{2}\left(4\right){\left(2.5\right)}^2 \\ =12.5\mathrm{m} \\ \mathrm{Distance}\mathrm{travelled}\mathrm{in}2.0\mathrm{s}: \\ {\mathrm{s}}_2=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ =\left(0\right)\left(2.0\right)+\frac{1}{2}\left(4\right){\left(2.0\right)}^2 \\ =8\mathrm{m} \\ \mathbf{Distance}\mathbf{}\mathbf{travelled}\mathbf{}\mathbf{in}\mathbf{}\mathbf{5}\mathbf{th}\mathbf{}\mathbf{half}\mathbf{}\mathbf{second}\mathbf{,}\mathbf{}\mathbf{that}\mathbf{}\mathbf{is}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{th}\mathbf{}\mathbf{second}\mathbf{:} \\ \mathrm{s}={\mathrm{s}}_1-{\mathrm{s}}_2 \\ =12.5-8 \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{m} \end{gathered}$$