Question Number 126475 by sdfg last updated on 20/Dec/20
Answered by physicstutes last updated on 20/Dec/20
![(a) a = ((𝚫v)/(𝚫t)) = (([(10 j) −(10 i + 20 j)] m s^(−1) )/((4−0) s)) = [−(((10)/4)) i − (((10)/4)) j ]m s^(−2) or a =( −(5/2)i − (5/2)j ) m s^(−2) ofcourse assuming we are using SI base units. (b) r = r_0 + v_0 t + (1/2)a t^2 but r_0 = 0^(→) since it is initially at the origin. ⇒ r = (10 i + 20 j)t + [(−(5/4) i −(5/4)j)]t^2 v = v_0 + at ⇒ v = (10 i + 20 j) + (−(5/2)i − (5/2)j)t](Q126479.png)
$$\:\left(\mathrm{a}\right)\:\boldsymbol{\mathrm{a}}\:=\:\frac{\boldsymbol{\Delta\mathrm{v}}}{\boldsymbol{\Delta}{t}}\:=\:\frac{\left[\left(\mathrm{10}\:\boldsymbol{\mathrm{j}}\right)\:−\left(\mathrm{10}\:\boldsymbol{\mathrm{i}}\:+\:\mathrm{20}\:\boldsymbol{\mathrm{j}}\right)\right]\:\mathrm{m}\:\mathrm{s}^{−\mathrm{1}} }{\left(\mathrm{4}−\mathrm{0}\right)\:\mathrm{s}}\:=\:\left[−\left(\frac{\mathrm{10}}{\mathrm{4}}\right)\:\boldsymbol{\mathrm{i}}\:−\:\left(\frac{\mathrm{10}}{\mathrm{4}}\right)\:\boldsymbol{\mathrm{j}}\:\right]\mathrm{m}\:\mathrm{s}^{−\mathrm{2}} \\ $$$$\mathrm{or}\:\boldsymbol{\mathrm{a}}\:=\left(\:−\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{i}}\:−\:\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{j}}\:\right)\:\mathrm{m}\:\mathrm{s}^{−\mathrm{2}} \:\:\:\mathrm{ofcourse}\:\mathrm{assuming}\:\mathrm{we}\:\mathrm{are}\:\mathrm{using}\:\mathrm{SI}\:\mathrm{base}\:\mathrm{units}. \\ $$$$\:\left(\mathrm{b}\right)\:\boldsymbol{\mathrm{r}}\:=\:\boldsymbol{\mathrm{r}}_{\mathrm{0}} +\:\:\boldsymbol{\mathrm{v}}_{\mathrm{0}} {t}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{a}}\:{t}^{\mathrm{2}} \:\mathrm{but}\:\boldsymbol{\mathrm{r}}_{\mathrm{0}} \:=\:\overset{\rightarrow} {\mathrm{0}}\:\mathrm{since}\:\mathrm{it}\:\mathrm{is}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}. \\ $$$$\Rightarrow\:\:\boldsymbol{\mathrm{r}}\:=\:\left(\mathrm{10}\:\boldsymbol{\mathrm{i}}\:+\:\mathrm{20}\:\boldsymbol{\mathrm{j}}\right){t}\:+\:\left[\left(−\frac{\mathrm{5}}{\mathrm{4}}\:\boldsymbol{\mathrm{i}}\:−\frac{\mathrm{5}}{\mathrm{4}}\boldsymbol{\mathrm{j}}\right)\right]{t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{v}}\:=\:\boldsymbol{\mathrm{v}}_{\mathrm{0}} \:+\:\boldsymbol{\mathrm{a}}{t}\: \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{v}}\:=\:\left(\mathrm{10}\:\boldsymbol{\mathrm{i}}\:+\:\mathrm{20}\:\boldsymbol{\mathrm{j}}\right)\:+\:\left(−\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{i}}\:−\:\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{j}}\right){t} \\ $$